Question

Ammonia is produced by the reaction of nitrogen and hydrogen according to the equation N2(g) + 3H2(g) → 2NH3(g) Calculate the mass of ammonia produced when 37.0 g of nitrogen react with 12.0 g of hydrogen. g NH3 Which is the excess reactant and how much of it will be left over when the reaction is complete? hydrogen nitrogen

Answer 1

The excess reactant is
`H_2`.

Left over = 1.9903 moles or 4.0122 g

Step-by-step explanation:

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Given: For
`N_2`

Given mass = 37.0 g

Molar mass of
`N_2` = 28.0134 g/mol

Moles of
`N_2` = 37.0 g / 28.0134 g/mol = 1.3208 moles

Given: For
`H_2`

Given mass = 12.0 g

Molar mass of
`H_2` = 2.0159 g/mol

Moles of
`H_2` = 12.0 g / 2.0159 g/mol = 5.9527 moles

According to the given reaction:

`N_2_((g))+3H_2_((g))\rightarrow 2NH_3_((g))`

1 mole of
`N_2` react with 3 moles of
`H_2`

1.3208 moles of
`N_2` react with 3*1.3208 moles of
`H_2`

Moles of
`H_2` reacted = 3.9624 moles

Available moles of
`H_2` = 5.9527 moles

Limiting reagent is the one which is present in small amount. Thus,
`N_2` is limiting reagent as
`H_2` is present in excess . (3.9624 < 5.9527)

The excess reactant is
`H_2`.

Left over = 5.9527 - 3.9624 moles = 1.9903 moles

Mass = moles × Molar mass = 1.9903 moles × 2.0159 g/mol = 4.0122 g

Left over is 4.0122 g of hydrogen.