Question

CH10,42C,H,OH +2CO, Upon decomposition of 0.12 mole sample of C6H,206, the mass of C,H,OH (M.m = 46 g/mol) obtained was 9.4 g. The percent yield of C,H,OH was:

a) 25%
b) 50%
c) 55%
d) 85%

Answer 1

To calculate the percent yield of C,H,OH, we need to use the given information about the mass of C,H,OH obtained and the molar mass of C6H,206. The percent yield is 50%.

Step-by-step explanation:

To calculate the percent yield of C,H,OH, we need to use the given information about the mass of C,H,OH obtained and the molar mass of C6H,206. First, let's calculate the number of moles of C,H,OH obtained: 9.4 g / 46 g/mol = 0.204 moles. Next, we can calculate the theoretical yield of C,H,OH by using the balanced chemical equation: C6H,206 → 10CO + 12H,O. From this equation, we can see that for each mole of C6H,206, we should obtain 10 moles of CO and 12 moles of H,O. Therefore, the theoretical yield of C,H,OH is 0.204 moles x 12 moles of H,O / 6 moles of C6H,206 = 0.408 moles. Finally, we can calculate the percent yield by dividing the actual yield (0.204 moles) by the theoretical yield (0.408 moles) and multiplying by 100. The percent yield is (0.204 moles / 0.408 moles) x 100 = 50%.