Question

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 h2 / 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 30 / 9 mol o2 x mol o2 / h2o x g h2o / h2o
limiting reactant= o2
excess reactant= h2

Explanation:

Step1: Calculate moles of H₂

$n_{H_2}=\frac{10.0g}{2.016g/mol}=4.96mol$

Step2: Use mole - ratio from balanced equation

For $2H_2 + O_2=2H_2O$, mole - ratio of $H_2$ to $H_2O$ is 1:1, so $n_{H_2O}=4.96mol$

Step3: Calculate mass of H₂O

$m_{H_2O}=n_{H_2O}\times M_{H_2O}=4.96mol\times18.015g/mol = 89.3g$

For O₂:

Step1: Calculate moles of O₂

$n_{O_2}=\frac{10.0g}{32.00g/mol}=0.3125mol$

Step2: Use mole - ratio from balanced equation

Mole - ratio of $O_2$ to $H_2O$ is 1:2, so $n_{H_2O}=0.3125mol\times2 = 0.625mol$

Step3: Calculate mass of H₂O

$m_{H_2O}=n_{H_2O}\times M_{H_2O}=0.625mol\times18.015g/mol = 11.3g$

Since O₂ produces less $H_2O$, O₂ is limiting reactant and H₂ is excess reactant.

Answer:

Limiting reactant: $O_2$
Excess reactant: $H_2$
Mass of $H_2O$ produced: $11.3g$