306991
For the reaction, \({\rm{2x + 3y + 4z}} \to {\rm{5w}}\) Intially if 1 mole of x, 3 moles of y and 4 moles of z is taken. If 1.25 moles of w is obtained then % yield of this reaction is
1 \(60\% \)
2 \(50\% \)
3 \(70\% \)
4 \(40\% \)
Explanation:
1 mole of x will give \(\frac{{\rm{5}}}{{\rm{2}}}{\rm{ = 2}}{\rm{.5}}{\mkern 1mu} {\mkern 1mu} {\rm{mol}}\) But, yield \({\rm{ = }}\frac{{{\rm{1}}{\rm{.25}}}}{{{\rm{2}}{\rm{.5}}}}{\rm{ \times 100 = 50\% }}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306992
How many moles of \(\mathrm{Zn}\left(\mathrm{FeS}_{2}\right)\) can be made from 2 moles of zinc, 3 moles of iron, and 5 moles of sulphur?
1 2 moles
2 3 moles
3 4 moles
4 5 moles
Explanation:
1 mole of \(\mathrm{Zn}\left(\mathrm{FeS}_{2}\right)\) contains: 1 mole of Zn , 1 mole of \(\mathrm{Fe}, 2\) moles of S . So, for 2 moles of \(\mathrm{Zn}\left(\mathrm{FeS}_{2}\right)\), we need 2 moles of \(\mathrm{Zn}, 2\) moles of Fe , and 4 moles of S . So maximum 2 moles of \(\mathrm{Zn}\left(\mathrm{FeS}_{2}\right)\) can be made. (1mole 'Fe' 1mole ' S ' will be unreacted).
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306993
0.4 g of dihydrogen is made to react with 7.1 g of dichlorine to form hydrogen chloride. The volume of hydrogen chloride formed at 273 K and 1 bar pressure is
\({\rm{2KCl}}{{\rm{O}}_{\rm{3}}}{\rm{ = 2 \times 122}}{\rm{.5}}{\mkern 1mu} {\mkern 1mu} {\rm{g}}\) 48 g of oxygen will be produced from 122.5 g of \({\rm{KCl}}{{\rm{O}}_{\rm{3}}}\) \(\therefore \) Amount of \({\rm{80\% KCl}}{{\rm{O}}_{\rm{3}}}\) needed \({\rm{ = }}\frac{{{\rm{100}}}}{{{\rm{80}}}}{\rm{ \times 122}}{\rm{.5 = 153}}{\rm{.12g}}\)
306991
For the reaction, \({\rm{2x + 3y + 4z}} \to {\rm{5w}}\) Intially if 1 mole of x, 3 moles of y and 4 moles of z is taken. If 1.25 moles of w is obtained then % yield of this reaction is
1 \(60\% \)
2 \(50\% \)
3 \(70\% \)
4 \(40\% \)
Explanation:
1 mole of x will give \(\frac{{\rm{5}}}{{\rm{2}}}{\rm{ = 2}}{\rm{.5}}{\mkern 1mu} {\mkern 1mu} {\rm{mol}}\) But, yield \({\rm{ = }}\frac{{{\rm{1}}{\rm{.25}}}}{{{\rm{2}}{\rm{.5}}}}{\rm{ \times 100 = 50\% }}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306992
How many moles of \(\mathrm{Zn}\left(\mathrm{FeS}_{2}\right)\) can be made from 2 moles of zinc, 3 moles of iron, and 5 moles of sulphur?
1 2 moles
2 3 moles
3 4 moles
4 5 moles
Explanation:
1 mole of \(\mathrm{Zn}\left(\mathrm{FeS}_{2}\right)\) contains: 1 mole of Zn , 1 mole of \(\mathrm{Fe}, 2\) moles of S . So, for 2 moles of \(\mathrm{Zn}\left(\mathrm{FeS}_{2}\right)\), we need 2 moles of \(\mathrm{Zn}, 2\) moles of Fe , and 4 moles of S . So maximum 2 moles of \(\mathrm{Zn}\left(\mathrm{FeS}_{2}\right)\) can be made. (1mole 'Fe' 1mole ' S ' will be unreacted).
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306993
0.4 g of dihydrogen is made to react with 7.1 g of dichlorine to form hydrogen chloride. The volume of hydrogen chloride formed at 273 K and 1 bar pressure is
\({\rm{2KCl}}{{\rm{O}}_{\rm{3}}}{\rm{ = 2 \times 122}}{\rm{.5}}{\mkern 1mu} {\mkern 1mu} {\rm{g}}\) 48 g of oxygen will be produced from 122.5 g of \({\rm{KCl}}{{\rm{O}}_{\rm{3}}}\) \(\therefore \) Amount of \({\rm{80\% KCl}}{{\rm{O}}_{\rm{3}}}\) needed \({\rm{ = }}\frac{{{\rm{100}}}}{{{\rm{80}}}}{\rm{ \times 122}}{\rm{.5 = 153}}{\rm{.12g}}\)
306991
For the reaction, \({\rm{2x + 3y + 4z}} \to {\rm{5w}}\) Intially if 1 mole of x, 3 moles of y and 4 moles of z is taken. If 1.25 moles of w is obtained then % yield of this reaction is
1 \(60\% \)
2 \(50\% \)
3 \(70\% \)
4 \(40\% \)
Explanation:
1 mole of x will give \(\frac{{\rm{5}}}{{\rm{2}}}{\rm{ = 2}}{\rm{.5}}{\mkern 1mu} {\mkern 1mu} {\rm{mol}}\) But, yield \({\rm{ = }}\frac{{{\rm{1}}{\rm{.25}}}}{{{\rm{2}}{\rm{.5}}}}{\rm{ \times 100 = 50\% }}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306992
How many moles of \(\mathrm{Zn}\left(\mathrm{FeS}_{2}\right)\) can be made from 2 moles of zinc, 3 moles of iron, and 5 moles of sulphur?
1 2 moles
2 3 moles
3 4 moles
4 5 moles
Explanation:
1 mole of \(\mathrm{Zn}\left(\mathrm{FeS}_{2}\right)\) contains: 1 mole of Zn , 1 mole of \(\mathrm{Fe}, 2\) moles of S . So, for 2 moles of \(\mathrm{Zn}\left(\mathrm{FeS}_{2}\right)\), we need 2 moles of \(\mathrm{Zn}, 2\) moles of Fe , and 4 moles of S . So maximum 2 moles of \(\mathrm{Zn}\left(\mathrm{FeS}_{2}\right)\) can be made. (1mole 'Fe' 1mole ' S ' will be unreacted).
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306993
0.4 g of dihydrogen is made to react with 7.1 g of dichlorine to form hydrogen chloride. The volume of hydrogen chloride formed at 273 K and 1 bar pressure is
\({\rm{2KCl}}{{\rm{O}}_{\rm{3}}}{\rm{ = 2 \times 122}}{\rm{.5}}{\mkern 1mu} {\mkern 1mu} {\rm{g}}\) 48 g of oxygen will be produced from 122.5 g of \({\rm{KCl}}{{\rm{O}}_{\rm{3}}}\) \(\therefore \) Amount of \({\rm{80\% KCl}}{{\rm{O}}_{\rm{3}}}\) needed \({\rm{ = }}\frac{{{\rm{100}}}}{{{\rm{80}}}}{\rm{ \times 122}}{\rm{.5 = 153}}{\rm{.12g}}\)
306991
For the reaction, \({\rm{2x + 3y + 4z}} \to {\rm{5w}}\) Intially if 1 mole of x, 3 moles of y and 4 moles of z is taken. If 1.25 moles of w is obtained then % yield of this reaction is
1 \(60\% \)
2 \(50\% \)
3 \(70\% \)
4 \(40\% \)
Explanation:
1 mole of x will give \(\frac{{\rm{5}}}{{\rm{2}}}{\rm{ = 2}}{\rm{.5}}{\mkern 1mu} {\mkern 1mu} {\rm{mol}}\) But, yield \({\rm{ = }}\frac{{{\rm{1}}{\rm{.25}}}}{{{\rm{2}}{\rm{.5}}}}{\rm{ \times 100 = 50\% }}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306992
How many moles of \(\mathrm{Zn}\left(\mathrm{FeS}_{2}\right)\) can be made from 2 moles of zinc, 3 moles of iron, and 5 moles of sulphur?
1 2 moles
2 3 moles
3 4 moles
4 5 moles
Explanation:
1 mole of \(\mathrm{Zn}\left(\mathrm{FeS}_{2}\right)\) contains: 1 mole of Zn , 1 mole of \(\mathrm{Fe}, 2\) moles of S . So, for 2 moles of \(\mathrm{Zn}\left(\mathrm{FeS}_{2}\right)\), we need 2 moles of \(\mathrm{Zn}, 2\) moles of Fe , and 4 moles of S . So maximum 2 moles of \(\mathrm{Zn}\left(\mathrm{FeS}_{2}\right)\) can be made. (1mole 'Fe' 1mole ' S ' will be unreacted).
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306993
0.4 g of dihydrogen is made to react with 7.1 g of dichlorine to form hydrogen chloride. The volume of hydrogen chloride formed at 273 K and 1 bar pressure is
\({\rm{2KCl}}{{\rm{O}}_{\rm{3}}}{\rm{ = 2 \times 122}}{\rm{.5}}{\mkern 1mu} {\mkern 1mu} {\rm{g}}\) 48 g of oxygen will be produced from 122.5 g of \({\rm{KCl}}{{\rm{O}}_{\rm{3}}}\) \(\therefore \) Amount of \({\rm{80\% KCl}}{{\rm{O}}_{\rm{3}}}\) needed \({\rm{ = }}\frac{{{\rm{100}}}}{{{\rm{80}}}}{\rm{ \times 122}}{\rm{.5 = 153}}{\rm{.12g}}\)
