Question

2H2 (1) + O2(g) → 2H20 (g)

1. Find the limiting reactant if you start with 30.0 grams of hydrogen and 5.29 grams of oxygen.

2. The actual yield for H2O in the above reaction is 6.72 g, Determine the percent yield for the reaction

when 9.93 grams of hydrogen and excess oxygen react?

Answer 1

Answer: 1. Oxygen is the the limiting reactant.

2. 7.52%

Step-by-step 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=(30.0g)/(2g/mol)=15.0moles`

`\text{Moles of} O_2=(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=
`(2)/(1)* 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=(9.93g)/(2g/mol)=4.96moles`

`\text{Moles of} H_2O=(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 =
`(2)/(2)* 4.96=4.96moles` of
`H_2O`

percentage yield =
`\frac{\text {actual yield}}{\text {theoretical yield}}=(0.373)/(4.96)* 100=7.52\%`

Thus the percent yield for the reaction is 7.52%