Question

If 0.0987 g H2O hydrolyzes, what is the theoretical yield of H2? 2H2O → → 2H2 + O2

Answer 1

The theoretical yield is 0.0110 g H₂.

Moles of H₂O = 0.0987 g H₂O × (1 mol H₂O/18.02 g H₂O) = 0.005 477 mol H₂O

Moles of H₂ = 0.005 477 mol H₂O × (2 mol H₂/2 mol H₂O) = 0.005 477 mol H₂

Theoretical yield of H₂ = 0.005 477 mol H₂ × (2.016 g H₂/1 mol H₂)

= 0.0110 g H₂

Answer 2

Answer: Theoretical yield of $H_2$ is 0.01 grams.

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to the molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

$\text{Number of moles of}H_2O=\frac{0.0987g}{18g/mol}=0.005moles$

$2H_2O\rightarrow 2H_2+O_2$

According to stoichiometry:

2 moles of water produces = 2 moles of hydrogen

Thus 0.005 moles of water produces =$\frac{2}{2}\times 0.005=0.005$ moles of hydrogen

Mass of $H_2$ produced =$moles\times {\text {molar mass}}}=0.005\times 2g/mol=0.01g$

Thus theoretical yield of $H_2$ is 0.01 grams.