Question

problem #1: for the combustion of sucrose: limiting reagent
h2 + o2 --> 2 h2o
there are 10.0 g of hydrogen gas (h2) and 10.0 g of oxygen gas (o2) reacting. which is the limiting reagent and excess reagent?
molar mass of h2o: h = 1.008 o = 16.00
∑(16 + 12.8) g/mol h2o

Explanation:

Step1: Calculate moles of reactants

$n_{H_2}=\frac{10.0\ g}{2.016\ g/mol}=4.96\ mol$, $n_{O_2}=\frac{10.0\ g}{32.00\ g/mol}=0.3125\ mol$

Step2: Use mole - ratio from equation

From $2H_2 + O_2
ightarrow2H_2O$, mole - ratio of $H_2$ to $O_2$ is 2:1. Moles of $H_2$ needed for 0.3125 mol of $O_2$ is $2\times0.3125 = 0.625\ mol$. Since $4.96\ mol>0.625\ mol$, $O_2$ is limiting.

Answer:

$O_2$ is the limiting reagent.