Answer to Question #183177 in General Chemistry for Ayesha

Question

Verify that it does not matter which product you use to predict the limiting reagent by using both products in this combustion reaction to determine the limiting reagent and the amount of the reactant in excess. Initial amounts of each reactant are given.

C3H8 (26.3 g) + 5O2 (21.8 g) →3CO2 (g) + 4H2O (l)

Answer 1

$26.3g C_3H_8×\frac{1molC_3H_8}{44.1gC_3H_8}×\frac{4molH_2O}{1molC_3H_8}×\frac{18gH_2O}{1molH_2O} = 43.03 mol$

$21.8gO_2×\frac{1molO_2}{32gO_2}×\frac{4molH_2O}{5molO_2}×\frac{18gH_2O}{1molH_2O} = 9.81 mol$

$\therefore O_2$ is limiting reagent

Amount of excess reagent left

$21.8gO_2×\frac{1molO_2}{32gO_2}×\frac{1molC_3H_8}{5molO_2}×\frac{44.1gC_3H_8}{1molC_3H_8} = 6.008 g C_3H_8$

$26.3gC_3H_8 - 6.008gC_3H_8 = 20.292 g C_3H_8 remain$