If 4.750 g of CO2(g) was obtained from the combustion of 1.638 g of propane, what was the percent yield?

To solve this problem, we need to follow these steps:

1. Write down the balanced chemical equation for the combustion of propane (C3H8). It is: C3H8 + 5O2 -> 3CO2 + 4H2O

2. From the balanced equation, we can see that one mole of propane (C3H8) produces three moles of carbon dioxide (CO2).

3. Calculate the molar mass of propane (C3H8). The molar mass of carbon (C) is approximately 12.01 g/mol, and the molar mass of hydrogen (H) is approximately 1.01 g/mol. Therefore, the molar mass of propane is (312.01 g/mol) + (81.01 g/mol) = 44.11 g/mol.

4. Calculate the number of moles of propane that were combusted. This is done by dividing the mass of propane that was combusted by the molar mass of propane. So, (1.638 g) / (44.11 g/mol) = 0.0371 mol of propane.

5. Calculate the number of moles of CO2 that should have been produced. According to the balanced chemical equation, one mole of propane produces three moles of CO2. Therefore, (0.0371 mol of propane) * (3 mol CO2 / 1 mol propane) = 0.1113 mol of CO2.

6. Calculate the mass of CO2 that should have been produced. The molar mass of CO2 is (112.01 g/mol) + (216.00 g/mol) = 44.01 g/mol. Therefore, (0.1113 mol of CO2) * (44.01 g/mol) = 4.898 g of CO2.

7. Calculate the percent yield. This is done by dividing the actual yield (the mass of CO2 that was actually produced) by the theoretical yield (the mass of CO2 that should have been produced), and then multiplying by 100%. So, (4.750 g / 4.898 g) * 100% = 96.98%.

Therefore, the percent yield of the reaction was approximately 96.98%.