Question

warm up limiting reagent
show your solution/ work (no work / solution = zero). observe correct number of significant figures. encircle or box your final answer.
problem #1: for the combustion of sucrose: 2 h₂ + 1 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 = 4.96 mol
o = 0.3125 mol
4.58 g/mol h₂o
1.008
x 0.3125 = 0.625 h₂ x 2 = 2.016
g h₂ to mol h₂ using atomic mass of h₂
mol h₂ to mol h₂o using balanced eq
mol h₂o to g h₂o using molar mass of h₂o
? g h₂o = 10.0 g h₂ x 1 mol h₂o / 16.128 h₂ x 2 mol h₂ / 2 h₂o x 84.68 g h₂o / 1 mol h₂o
g o₂ to mol o₂ using atomic mass of o₂
mol o₂ to mol h₂o using balanced eq
mol h₂o to g h₂o using molar mass of h₂o
? g h₂o = 10.0 g o₂ x 3 o₂ / 1 mol o₂ x mol o₂ / h₂o x g h₂o / h₂o
balanced eq
limiting reactant = o₂
excess reactant = h₂

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 balanced equation

From $2H_2 + O_2
ightarrow2H_2O$, mole - ratio of $H_2$ to $O_2$ is 2:1. For 0.3125 mol of $O_2$, moles of $H_2$ needed is $2\times0.3125 = 0.625\ mol$. 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: $H_2$