Question

2H2 (1) + O2(g) → 2H20 (g)

Answer 1

Answer: 1. Oxygen is the the limiting reactant.

2. 7.52%

Explanation:

The balanced chemical equation is:

$2H_2(g)+O_2(g)\rightarrow 2H_2O(g)$  

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$    

$\text{Moles of} H_2=\frac{30.0g}{2g/mol}=15.0moles$

$\text{Moles of} O_2=\frac{5.29g}{32g/mol}=0.165moles$

According to stoichiometry :

1 mole of $O_2$ require  = 2 moles of $H_2$

Thus 0.165 moles of $O_2$ will require=$\frac{2}{1}\times 0.165=0.331moles$  of $NH_3$

Thus $O_2$ is the limiting reagent as it limits the formation of product and $H_2$ is the excess reagent.

2.  $\text{Moles of} H_2=\frac{9.93g}{2g/mol}=4.96moles$

$\text{Moles of} H_2O=\frac{6.72g}{18g/mol}=0.373moles$

As 2 moles of $H_2$ give = 2 moles of $H_2O$

Thus 4.96 moles of $H_2$ give =$\frac{2}{2}\times 4.96=4.96moles$  of $H_2O$

percentage yield = $\frac{\text {actual yield}}{\text {theoretical yield}}=\frac{0.373}{4.96}\times 100=7.52\%$

Thus the percent yield for the reaction is 7.52%