228699
Calculate the amount of hydrogen gas required in order to produce $100 \mathrm{~g}$ of ammonia by the reaction of $\mathrm{N}_2(\mathrm{~g})$ and $\mathrm{H}_2(\mathrm{~g})$ gases.
1 $35.29 \mathrm{~g}$
2 $17.65 \mathrm{~g}$
3 $28.11 \mathrm{~g}$
4 $34 \mathrm{~g}$ AP
Explanation:
Given, $\mathrm{NH}_3$ Produce $100 \mathrm{gm}$. $\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_3$ So, the number of moles of $\begin{aligned} \mathrm{NH}_3 & =\frac{\text { Weight }}{\text { Molecular weight }} \\ \mathrm{n}_{\mathrm{NH}_3} & =\frac{100}{17} \\ \mathrm{n}_{\mathrm{NH}_3} & =5.88 \text { mole. } \end{aligned}$ So, the amount of hydrogen gas Required is $=3 \times 5.88=17.65 \mathrm{gm}$.
AP EAPCET 24.08.2021 Shift-II
Some Basic Concepts of Chemistry
228700
When 20 g of $\mathrm{CaCO}_3$ is treated with $20 \mathrm{~g}$ of $\mathrm{HCl}$ the mass of $\mathrm{CO}_2$ formed would be
1 $10 \mathrm{~g}$
2 $8.8 \mathrm{~g}$
3 $22.2 \mathrm{~g}$
4 $20 \mathrm{~g}$ Ans.
Explanation:
$\begin{array}{ccc} \mathrm{CaCO}_3(\mathrm{~s})&+&2 \mathrm{HCl}(\mathrm{aq}) & \rightarrow& \mathrm{CaCl}_2(\mathrm{aq})&+&\mathrm{H}_2 \mathrm{O}(l)&+&\mathrm{CO}_2(\mathrm{~g}) \\ 1 \mathrm{~mol} & &2 \mathrm{~mol} & &&&&&1 \mathrm{~mol} \\ 100 \mathrm{~g} & &73 \mathrm{~g} & &&&&&44 \mathrm{~g} \end{array}$ Let $\mathrm{CaCO}_3(\mathrm{~s})$ be completely consumed in the reaction. $\therefore 100 \mathrm{~g} \mathrm{CaCO}_3$ give $44 \mathrm{~g} \mathrm{CO}_2$ $\therefore 20 \mathrm{~g} \mathrm{CaCO}_3$ will give $\frac{44}{100} \times 20 \mathrm{~g} \mathrm{CO}_2=8.8 \mathrm{~g} \mathrm{CO}_2$
AP EAPCET 19-08-2021 Shift-I
Some Basic Concepts of Chemistry
228701
The complete combustion of one mole of benzene produces grams of carbon dioxide.
228705
The minimum amount of $\mathrm{O}_2(\mathrm{~g})$ consumed per gram of reactant is for the reaction (Given atomic mass : $F e=56, O=16, \mathrm{Mg}=24, \mathrm{P}=31$, ) $\mathrm{C}=12, \mathrm{H}=1$ )
228706
$25 \mathrm{~g}$ of an unknown hydrocarbon upon burning produces 88 g of $\mathrm{CO}_2$ and $9 \mathrm{~g}$ of $\mathrm{H}_2 \mathrm{O}$. This unknown hydrocarbon contains
1 $20 \mathrm{~g}$ of carbon and $5 \mathrm{~g}$ of hydrogen
2 $22 \mathrm{~g}$ of carbon and $3 \mathrm{~g}$ of hydrogen
3 $24 \mathrm{~g}$ of carbon and $1 \mathrm{~g}$ of hydrogen
4 $18 \mathrm{~g}$ of carbon and $7 \mathrm{~g}$ of hydrogen
Explanation:
Let the hydrocarbon is $\mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{y}}$. $\mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{y}}+\mathrm{O}_2 \longrightarrow \mathrm{xCO}_2+\frac{\mathrm{y}}{2} \mathrm{H}_2 \mathrm{O}$ The weight of Carbon $=\frac{88}{44} \times 12$ $=24 \mathrm{~g}$ The weight of hydrogen $=\frac{9}{18} \times 1$ $\begin{aligned} & \quad=24 \mathrm{~g} \\ & \mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{y}}+\left(\mathrm{x}+\frac{\mathrm{y}}{4}\right) \mathrm{O}_2 \rightarrow \mathrm{xCO}_2+\frac{\mathrm{y}}{2} \mathrm{H}_2 \mathrm{O} \\ & \mathrm{C}=\mathrm{x} \times \frac{25}{\mathrm{M}}=2 \\ & \mathrm{H}=\frac{\mathrm{y}}{2} \times \frac{25}{\mathrm{M}}=1 \\ & \Rightarrow 24: 1 \text { ratio by mass } \end{aligned}$
228699
Calculate the amount of hydrogen gas required in order to produce $100 \mathrm{~g}$ of ammonia by the reaction of $\mathrm{N}_2(\mathrm{~g})$ and $\mathrm{H}_2(\mathrm{~g})$ gases.
1 $35.29 \mathrm{~g}$
2 $17.65 \mathrm{~g}$
3 $28.11 \mathrm{~g}$
4 $34 \mathrm{~g}$ AP
Explanation:
Given, $\mathrm{NH}_3$ Produce $100 \mathrm{gm}$. $\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_3$ So, the number of moles of $\begin{aligned} \mathrm{NH}_3 & =\frac{\text { Weight }}{\text { Molecular weight }} \\ \mathrm{n}_{\mathrm{NH}_3} & =\frac{100}{17} \\ \mathrm{n}_{\mathrm{NH}_3} & =5.88 \text { mole. } \end{aligned}$ So, the amount of hydrogen gas Required is $=3 \times 5.88=17.65 \mathrm{gm}$.
AP EAPCET 24.08.2021 Shift-II
Some Basic Concepts of Chemistry
228700
When 20 g of $\mathrm{CaCO}_3$ is treated with $20 \mathrm{~g}$ of $\mathrm{HCl}$ the mass of $\mathrm{CO}_2$ formed would be
1 $10 \mathrm{~g}$
2 $8.8 \mathrm{~g}$
3 $22.2 \mathrm{~g}$
4 $20 \mathrm{~g}$ Ans.
Explanation:
$\begin{array}{ccc} \mathrm{CaCO}_3(\mathrm{~s})&+&2 \mathrm{HCl}(\mathrm{aq}) & \rightarrow& \mathrm{CaCl}_2(\mathrm{aq})&+&\mathrm{H}_2 \mathrm{O}(l)&+&\mathrm{CO}_2(\mathrm{~g}) \\ 1 \mathrm{~mol} & &2 \mathrm{~mol} & &&&&&1 \mathrm{~mol} \\ 100 \mathrm{~g} & &73 \mathrm{~g} & &&&&&44 \mathrm{~g} \end{array}$ Let $\mathrm{CaCO}_3(\mathrm{~s})$ be completely consumed in the reaction. $\therefore 100 \mathrm{~g} \mathrm{CaCO}_3$ give $44 \mathrm{~g} \mathrm{CO}_2$ $\therefore 20 \mathrm{~g} \mathrm{CaCO}_3$ will give $\frac{44}{100} \times 20 \mathrm{~g} \mathrm{CO}_2=8.8 \mathrm{~g} \mathrm{CO}_2$
AP EAPCET 19-08-2021 Shift-I
Some Basic Concepts of Chemistry
228701
The complete combustion of one mole of benzene produces grams of carbon dioxide.
228705
The minimum amount of $\mathrm{O}_2(\mathrm{~g})$ consumed per gram of reactant is for the reaction (Given atomic mass : $F e=56, O=16, \mathrm{Mg}=24, \mathrm{P}=31$, ) $\mathrm{C}=12, \mathrm{H}=1$ )
228706
$25 \mathrm{~g}$ of an unknown hydrocarbon upon burning produces 88 g of $\mathrm{CO}_2$ and $9 \mathrm{~g}$ of $\mathrm{H}_2 \mathrm{O}$. This unknown hydrocarbon contains
1 $20 \mathrm{~g}$ of carbon and $5 \mathrm{~g}$ of hydrogen
2 $22 \mathrm{~g}$ of carbon and $3 \mathrm{~g}$ of hydrogen
3 $24 \mathrm{~g}$ of carbon and $1 \mathrm{~g}$ of hydrogen
4 $18 \mathrm{~g}$ of carbon and $7 \mathrm{~g}$ of hydrogen
Explanation:
Let the hydrocarbon is $\mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{y}}$. $\mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{y}}+\mathrm{O}_2 \longrightarrow \mathrm{xCO}_2+\frac{\mathrm{y}}{2} \mathrm{H}_2 \mathrm{O}$ The weight of Carbon $=\frac{88}{44} \times 12$ $=24 \mathrm{~g}$ The weight of hydrogen $=\frac{9}{18} \times 1$ $\begin{aligned} & \quad=24 \mathrm{~g} \\ & \mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{y}}+\left(\mathrm{x}+\frac{\mathrm{y}}{4}\right) \mathrm{O}_2 \rightarrow \mathrm{xCO}_2+\frac{\mathrm{y}}{2} \mathrm{H}_2 \mathrm{O} \\ & \mathrm{C}=\mathrm{x} \times \frac{25}{\mathrm{M}}=2 \\ & \mathrm{H}=\frac{\mathrm{y}}{2} \times \frac{25}{\mathrm{M}}=1 \\ & \Rightarrow 24: 1 \text { ratio by mass } \end{aligned}$
228699
Calculate the amount of hydrogen gas required in order to produce $100 \mathrm{~g}$ of ammonia by the reaction of $\mathrm{N}_2(\mathrm{~g})$ and $\mathrm{H}_2(\mathrm{~g})$ gases.
1 $35.29 \mathrm{~g}$
2 $17.65 \mathrm{~g}$
3 $28.11 \mathrm{~g}$
4 $34 \mathrm{~g}$ AP
Explanation:
Given, $\mathrm{NH}_3$ Produce $100 \mathrm{gm}$. $\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_3$ So, the number of moles of $\begin{aligned} \mathrm{NH}_3 & =\frac{\text { Weight }}{\text { Molecular weight }} \\ \mathrm{n}_{\mathrm{NH}_3} & =\frac{100}{17} \\ \mathrm{n}_{\mathrm{NH}_3} & =5.88 \text { mole. } \end{aligned}$ So, the amount of hydrogen gas Required is $=3 \times 5.88=17.65 \mathrm{gm}$.
AP EAPCET 24.08.2021 Shift-II
Some Basic Concepts of Chemistry
228700
When 20 g of $\mathrm{CaCO}_3$ is treated with $20 \mathrm{~g}$ of $\mathrm{HCl}$ the mass of $\mathrm{CO}_2$ formed would be
1 $10 \mathrm{~g}$
2 $8.8 \mathrm{~g}$
3 $22.2 \mathrm{~g}$
4 $20 \mathrm{~g}$ Ans.
Explanation:
$\begin{array}{ccc} \mathrm{CaCO}_3(\mathrm{~s})&+&2 \mathrm{HCl}(\mathrm{aq}) & \rightarrow& \mathrm{CaCl}_2(\mathrm{aq})&+&\mathrm{H}_2 \mathrm{O}(l)&+&\mathrm{CO}_2(\mathrm{~g}) \\ 1 \mathrm{~mol} & &2 \mathrm{~mol} & &&&&&1 \mathrm{~mol} \\ 100 \mathrm{~g} & &73 \mathrm{~g} & &&&&&44 \mathrm{~g} \end{array}$ Let $\mathrm{CaCO}_3(\mathrm{~s})$ be completely consumed in the reaction. $\therefore 100 \mathrm{~g} \mathrm{CaCO}_3$ give $44 \mathrm{~g} \mathrm{CO}_2$ $\therefore 20 \mathrm{~g} \mathrm{CaCO}_3$ will give $\frac{44}{100} \times 20 \mathrm{~g} \mathrm{CO}_2=8.8 \mathrm{~g} \mathrm{CO}_2$
AP EAPCET 19-08-2021 Shift-I
Some Basic Concepts of Chemistry
228701
The complete combustion of one mole of benzene produces grams of carbon dioxide.
228705
The minimum amount of $\mathrm{O}_2(\mathrm{~g})$ consumed per gram of reactant is for the reaction (Given atomic mass : $F e=56, O=16, \mathrm{Mg}=24, \mathrm{P}=31$, ) $\mathrm{C}=12, \mathrm{H}=1$ )
228706
$25 \mathrm{~g}$ of an unknown hydrocarbon upon burning produces 88 g of $\mathrm{CO}_2$ and $9 \mathrm{~g}$ of $\mathrm{H}_2 \mathrm{O}$. This unknown hydrocarbon contains
1 $20 \mathrm{~g}$ of carbon and $5 \mathrm{~g}$ of hydrogen
2 $22 \mathrm{~g}$ of carbon and $3 \mathrm{~g}$ of hydrogen
3 $24 \mathrm{~g}$ of carbon and $1 \mathrm{~g}$ of hydrogen
4 $18 \mathrm{~g}$ of carbon and $7 \mathrm{~g}$ of hydrogen
Explanation:
Let the hydrocarbon is $\mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{y}}$. $\mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{y}}+\mathrm{O}_2 \longrightarrow \mathrm{xCO}_2+\frac{\mathrm{y}}{2} \mathrm{H}_2 \mathrm{O}$ The weight of Carbon $=\frac{88}{44} \times 12$ $=24 \mathrm{~g}$ The weight of hydrogen $=\frac{9}{18} \times 1$ $\begin{aligned} & \quad=24 \mathrm{~g} \\ & \mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{y}}+\left(\mathrm{x}+\frac{\mathrm{y}}{4}\right) \mathrm{O}_2 \rightarrow \mathrm{xCO}_2+\frac{\mathrm{y}}{2} \mathrm{H}_2 \mathrm{O} \\ & \mathrm{C}=\mathrm{x} \times \frac{25}{\mathrm{M}}=2 \\ & \mathrm{H}=\frac{\mathrm{y}}{2} \times \frac{25}{\mathrm{M}}=1 \\ & \Rightarrow 24: 1 \text { ratio by mass } \end{aligned}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Some Basic Concepts of Chemistry
228699
Calculate the amount of hydrogen gas required in order to produce $100 \mathrm{~g}$ of ammonia by the reaction of $\mathrm{N}_2(\mathrm{~g})$ and $\mathrm{H}_2(\mathrm{~g})$ gases.
1 $35.29 \mathrm{~g}$
2 $17.65 \mathrm{~g}$
3 $28.11 \mathrm{~g}$
4 $34 \mathrm{~g}$ AP
Explanation:
Given, $\mathrm{NH}_3$ Produce $100 \mathrm{gm}$. $\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_3$ So, the number of moles of $\begin{aligned} \mathrm{NH}_3 & =\frac{\text { Weight }}{\text { Molecular weight }} \\ \mathrm{n}_{\mathrm{NH}_3} & =\frac{100}{17} \\ \mathrm{n}_{\mathrm{NH}_3} & =5.88 \text { mole. } \end{aligned}$ So, the amount of hydrogen gas Required is $=3 \times 5.88=17.65 \mathrm{gm}$.
AP EAPCET 24.08.2021 Shift-II
Some Basic Concepts of Chemistry
228700
When 20 g of $\mathrm{CaCO}_3$ is treated with $20 \mathrm{~g}$ of $\mathrm{HCl}$ the mass of $\mathrm{CO}_2$ formed would be
1 $10 \mathrm{~g}$
2 $8.8 \mathrm{~g}$
3 $22.2 \mathrm{~g}$
4 $20 \mathrm{~g}$ Ans.
Explanation:
$\begin{array}{ccc} \mathrm{CaCO}_3(\mathrm{~s})&+&2 \mathrm{HCl}(\mathrm{aq}) & \rightarrow& \mathrm{CaCl}_2(\mathrm{aq})&+&\mathrm{H}_2 \mathrm{O}(l)&+&\mathrm{CO}_2(\mathrm{~g}) \\ 1 \mathrm{~mol} & &2 \mathrm{~mol} & &&&&&1 \mathrm{~mol} \\ 100 \mathrm{~g} & &73 \mathrm{~g} & &&&&&44 \mathrm{~g} \end{array}$ Let $\mathrm{CaCO}_3(\mathrm{~s})$ be completely consumed in the reaction. $\therefore 100 \mathrm{~g} \mathrm{CaCO}_3$ give $44 \mathrm{~g} \mathrm{CO}_2$ $\therefore 20 \mathrm{~g} \mathrm{CaCO}_3$ will give $\frac{44}{100} \times 20 \mathrm{~g} \mathrm{CO}_2=8.8 \mathrm{~g} \mathrm{CO}_2$
AP EAPCET 19-08-2021 Shift-I
Some Basic Concepts of Chemistry
228701
The complete combustion of one mole of benzene produces grams of carbon dioxide.
228705
The minimum amount of $\mathrm{O}_2(\mathrm{~g})$ consumed per gram of reactant is for the reaction (Given atomic mass : $F e=56, O=16, \mathrm{Mg}=24, \mathrm{P}=31$, ) $\mathrm{C}=12, \mathrm{H}=1$ )
228706
$25 \mathrm{~g}$ of an unknown hydrocarbon upon burning produces 88 g of $\mathrm{CO}_2$ and $9 \mathrm{~g}$ of $\mathrm{H}_2 \mathrm{O}$. This unknown hydrocarbon contains
1 $20 \mathrm{~g}$ of carbon and $5 \mathrm{~g}$ of hydrogen
2 $22 \mathrm{~g}$ of carbon and $3 \mathrm{~g}$ of hydrogen
3 $24 \mathrm{~g}$ of carbon and $1 \mathrm{~g}$ of hydrogen
4 $18 \mathrm{~g}$ of carbon and $7 \mathrm{~g}$ of hydrogen
Explanation:
Let the hydrocarbon is $\mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{y}}$. $\mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{y}}+\mathrm{O}_2 \longrightarrow \mathrm{xCO}_2+\frac{\mathrm{y}}{2} \mathrm{H}_2 \mathrm{O}$ The weight of Carbon $=\frac{88}{44} \times 12$ $=24 \mathrm{~g}$ The weight of hydrogen $=\frac{9}{18} \times 1$ $\begin{aligned} & \quad=24 \mathrm{~g} \\ & \mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{y}}+\left(\mathrm{x}+\frac{\mathrm{y}}{4}\right) \mathrm{O}_2 \rightarrow \mathrm{xCO}_2+\frac{\mathrm{y}}{2} \mathrm{H}_2 \mathrm{O} \\ & \mathrm{C}=\mathrm{x} \times \frac{25}{\mathrm{M}}=2 \\ & \mathrm{H}=\frac{\mathrm{y}}{2} \times \frac{25}{\mathrm{M}}=1 \\ & \Rightarrow 24: 1 \text { ratio by mass } \end{aligned}$
228699
Calculate the amount of hydrogen gas required in order to produce $100 \mathrm{~g}$ of ammonia by the reaction of $\mathrm{N}_2(\mathrm{~g})$ and $\mathrm{H}_2(\mathrm{~g})$ gases.
1 $35.29 \mathrm{~g}$
2 $17.65 \mathrm{~g}$
3 $28.11 \mathrm{~g}$
4 $34 \mathrm{~g}$ AP
Explanation:
Given, $\mathrm{NH}_3$ Produce $100 \mathrm{gm}$. $\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_3$ So, the number of moles of $\begin{aligned} \mathrm{NH}_3 & =\frac{\text { Weight }}{\text { Molecular weight }} \\ \mathrm{n}_{\mathrm{NH}_3} & =\frac{100}{17} \\ \mathrm{n}_{\mathrm{NH}_3} & =5.88 \text { mole. } \end{aligned}$ So, the amount of hydrogen gas Required is $=3 \times 5.88=17.65 \mathrm{gm}$.
AP EAPCET 24.08.2021 Shift-II
Some Basic Concepts of Chemistry
228700
When 20 g of $\mathrm{CaCO}_3$ is treated with $20 \mathrm{~g}$ of $\mathrm{HCl}$ the mass of $\mathrm{CO}_2$ formed would be
1 $10 \mathrm{~g}$
2 $8.8 \mathrm{~g}$
3 $22.2 \mathrm{~g}$
4 $20 \mathrm{~g}$ Ans.
Explanation:
$\begin{array}{ccc} \mathrm{CaCO}_3(\mathrm{~s})&+&2 \mathrm{HCl}(\mathrm{aq}) & \rightarrow& \mathrm{CaCl}_2(\mathrm{aq})&+&\mathrm{H}_2 \mathrm{O}(l)&+&\mathrm{CO}_2(\mathrm{~g}) \\ 1 \mathrm{~mol} & &2 \mathrm{~mol} & &&&&&1 \mathrm{~mol} \\ 100 \mathrm{~g} & &73 \mathrm{~g} & &&&&&44 \mathrm{~g} \end{array}$ Let $\mathrm{CaCO}_3(\mathrm{~s})$ be completely consumed in the reaction. $\therefore 100 \mathrm{~g} \mathrm{CaCO}_3$ give $44 \mathrm{~g} \mathrm{CO}_2$ $\therefore 20 \mathrm{~g} \mathrm{CaCO}_3$ will give $\frac{44}{100} \times 20 \mathrm{~g} \mathrm{CO}_2=8.8 \mathrm{~g} \mathrm{CO}_2$
AP EAPCET 19-08-2021 Shift-I
Some Basic Concepts of Chemistry
228701
The complete combustion of one mole of benzene produces grams of carbon dioxide.
228705
The minimum amount of $\mathrm{O}_2(\mathrm{~g})$ consumed per gram of reactant is for the reaction (Given atomic mass : $F e=56, O=16, \mathrm{Mg}=24, \mathrm{P}=31$, ) $\mathrm{C}=12, \mathrm{H}=1$ )
228706
$25 \mathrm{~g}$ of an unknown hydrocarbon upon burning produces 88 g of $\mathrm{CO}_2$ and $9 \mathrm{~g}$ of $\mathrm{H}_2 \mathrm{O}$. This unknown hydrocarbon contains
1 $20 \mathrm{~g}$ of carbon and $5 \mathrm{~g}$ of hydrogen
2 $22 \mathrm{~g}$ of carbon and $3 \mathrm{~g}$ of hydrogen
3 $24 \mathrm{~g}$ of carbon and $1 \mathrm{~g}$ of hydrogen
4 $18 \mathrm{~g}$ of carbon and $7 \mathrm{~g}$ of hydrogen
Explanation:
Let the hydrocarbon is $\mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{y}}$. $\mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{y}}+\mathrm{O}_2 \longrightarrow \mathrm{xCO}_2+\frac{\mathrm{y}}{2} \mathrm{H}_2 \mathrm{O}$ The weight of Carbon $=\frac{88}{44} \times 12$ $=24 \mathrm{~g}$ The weight of hydrogen $=\frac{9}{18} \times 1$ $\begin{aligned} & \quad=24 \mathrm{~g} \\ & \mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{y}}+\left(\mathrm{x}+\frac{\mathrm{y}}{4}\right) \mathrm{O}_2 \rightarrow \mathrm{xCO}_2+\frac{\mathrm{y}}{2} \mathrm{H}_2 \mathrm{O} \\ & \mathrm{C}=\mathrm{x} \times \frac{25}{\mathrm{M}}=2 \\ & \mathrm{H}=\frac{\mathrm{y}}{2} \times \frac{25}{\mathrm{M}}=1 \\ & \Rightarrow 24: 1 \text { ratio by mass } \end{aligned}$