Question

For the following reaction, 22.9 grams of nitrogen monoxide are allowed to react with 5.80 grams of hydrogen gas. nitrogen monoxide (g) + hydrogen (g)>nitrogen (g) + water (1) What is the maximum amount of nitrogen gas that can be formed? What is the FORMULA for the limiting reagent? grams What amount of the excess reagent remains after the reaction is complete? grams

Answer 1

1) Maximun ammount of nitrogen gas:
`m_(N2)=10.682 g N_2`

2) Limiting reagent:
`NO`

3) Ammount of excess reagent:
`m_(N2)=4.274 g`

Step-by-step explanation:

The reaction

`2 NO (g) + 2 H_2 (g) \longrightarrow N_2 (g) + 2 H_2O (g)`

Moles of nitrogen monoxide

Molecular weight:
`M_(NO)=30 g/mol`

`n_(NO)=(m_(NO))/(M_(NO))`

`n_(NO)=(22.9 g)/(30 g/mol)=0.763 mol`

Moles of hydrogen

Molecular weight:
`M_(H2)=2 g/mol`

`n_(H2)=(m_(H2))/(M_(H2))`

`n_(H2)=(.5.8 g)/(2 g/mol)=2.9 mol`

Mol rate of H2 and NO is 1:1 => hydrogen gas is in excess

1) Maximun ammount of nitrogen gas => when all NO reacted

`m_(N2)=0.763 mol NO* (1 mol N_2)/(2 mol NO)*(28 g N_2)/(mol N_2)`

`m_(N2)=10.682 g N_2`

2) Limiting reagent:
`NO`

3) Ammount of excess reagent:

`m_(N2)=(2.9 mol - 0.763 mol NO* (1 mol H_2)/(1 mol NO))*(2 g H_2)/(mol H_2)`

`m_(N2)=4.274 g`