Question

1. hydrogen gas and oxygen gas react to form water vapor. suppose you have 5.00 g of hydrogen and 13.0 g of oxygen in a reactor.

a. write the balanced equation.
b. which is the limiting reactant? show your molar mass calculations
c. how much of the excess reactant is left?
d. the reaction actually produces 10.45 g of water. what is the percent yield?

Explanation:

Step1: Write balanced equation

$$2H_{2}(g)+O_{2}(g) ightarrow 2H_{2}O(g)$$

Step2: Calculate moles of reactants

Molar mass of \(H_{2}\): \(M_{H_{2}} = 2\space g/mol\). Moles of \(H_{2}\), \(n_{H_{2}}=\frac{5.00\space g}{2\space g/mol}= 2.5\space mol\)
Molar mass of \(O_{2}\): \(M_{O_{2}} = 32\space g/mol\). Moles of \(O_{2}\), \(n_{O_{2}}=\frac{13.0\space g}{32\space g/mol}\approx0.406\space mol\)
From the balanced equation, mole ratio \(H_{2}:O_{2}=2:1\). For \(0.406\space mol\) of \(O_{2}\), moles of \(H_{2}\) required \(n_{H_{2}\text{(req)}} = 2\times0.406 = 0.812\space mol\). Since \(2.5\space mol>0.812\space mol\), \(O_{2}\) is the limiting reactant.

Step3: Calculate excess reactant left

Moles of \(H_{2}\) reacted \(= 0.812\space mol\). Moles of \(H_{2}\) left \(n_{H_{2}\text{(left)}}=2.5 - 0.812=1.688\space mol\). Mass of \(H_{2}\) left \(m_{H_{2}\text{(left)}}=1.688\space mol\times2\space g/mol = 3.376\space g\)

Step4: Calculate theoretical yield and percent yield

From balanced equation, mole ratio \(O_{2}:H_{2}O = 1:2\). Moles of \(H_{2}O\) formed \(n_{H_{2}O}=2\times0.406 = 0.812\space mol\). Molar mass of \(H_{2}O\), \(M_{H_{2}O}=18\space g/mol\). Theoretical yield \(m_{H_{2}O\text{(theo)}}=0.812\space mol\times18\space g/mol = 14.616\space g\)
Percent yield \(=\frac{10.45\space g}{14.616\space g}\times100\%\approx71.5\%\)

Answer:

a. \(2H_{2}(g)+O_{2}(g)
ightarrow 2H_{2}O(g)\)
b. \(O_{2}\) is the limiting reactant
c. \(3.376\space g\) of \(H_{2}\) is left
d. The percent yield is approximately \(71.5\%\)