Question

encircle or box your final answer. limiting reagent problem #1: for the combustion of sucrose: \\_\\_h₂ + \\_\\_o₂ --> 2h₂o 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 18.016 g/mol h₂o

Explanation:

Step1: Balance the equation

$2H_2 + O_2
ightarrow 2H_2O$

Step2: Calculate moles of reactants

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

Step3: Determine limiting reagent

From the equation, mole - ratio of $H_2$ to $O_2$ is 2:1. For 0.3125 mol of $O_2$, we need $2\times0.3125 = 0.625$ mol of $H_2$. Since we have 4.96 mol of $H_2$, $O_2$ is the limiting reagent and $H_2$ is the excess reagent.

Answer:

Limiting reagent: $O_2$, Excess reagent: $H_2$