Question

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

Answer 1

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

Step-by-step explanation:

According to avogadro's law, 1 mole of every substance weighs equal to the molecular mass and contains avogadro's number
`6.023* 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=(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 =
`(2)/(2)* 0.005=0.005` moles of hydrogen

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

Thus theoretical yield of
`H_2` is 0.01 grams.

Answer 2

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