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If 10.5 g of hydrogen, H2, were mixed with 6.51 g of acetylene, C2H2, and allowed to react according to the following equation,...Asked by C.M.
If 10.5 g of hydrogen,H2, were mixed with 6.51 g of acetylene, C2H2, and allowed to react according to the following equation, what is the theoretical yield of ethane, C2H6, produced? If 7.00 g of ethane, C2H6, were recovered at the end of the reaction, what is the percent yield?. C2H2(g)+2H2(g)=>C2H6(g)
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Answered by
Dorothy
Molar mass of acetylene is 35.01 g/mol
n=6.51/35.01= 0.1859 mol of C2H2
1 mol C2H2 0.1859 mol C2H2
__________= ________________
1 mol C2H6 X mol C2H6
x= 0.1859 mol C2H6
Molar mass of ethanol is 38.06 g/mol
m=(0.1859 mol)(38.06)= 7.07 g this is theoretical yield
% yield 7.00 g/7.07g x100%= 99% this is percentage yield
n=6.51/35.01= 0.1859 mol of C2H2
1 mol C2H2 0.1859 mol C2H2
__________= ________________
1 mol C2H6 X mol C2H6
x= 0.1859 mol C2H6
Molar mass of ethanol is 38.06 g/mol
m=(0.1859 mol)(38.06)= 7.07 g this is theoretical yield
% yield 7.00 g/7.07g x100%= 99% this is percentage yield
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