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

The excess reactant is .

Left over = 1.9903 moles or 4.0122 g

Explanation:

The formula for the calculation of moles is shown below:

Given: For

Given mass = 37.0 g

Molar mass of = 28.0134 g/mol

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

Given: For

Given mass = 12.0 g

Molar mass of = 2.0159 g/mol

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

According to the given reaction:

1 mole of react with 3 moles of

1.3208 moles of react with 3*1.3208 moles of

Moles of reacted = 3.9624 moles

Available moles of = 5.9527 moles

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

The excess reactant is .

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.

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"