# given the foloowing chemical e

given the foloowing chemical equation: H2+3N2 yeilds2NH3…….

if you react 26.5g of nitrogen gas with 13.67g of hydrogen gas,how many grams of ammonia gas are produced? which reactant islimiting? how many grams of the excess reactant remain?

Answer:

1)

Molar mass of H2 = 2.016 g/mol

mass of H2 = 13.67 g

we have below equation to be used:

number of mol of H2,

n = mass of H2/molar mass of H2

=(13.67 g)/(2.016 g/mol)

= 6.781 mol

Molar mass of N2 = 28.02 g/mol

mass of N2 = 26.5 g

we have below equation to be used:

number of mol of N2,

n = mass of N2/molar mass of N2

=(26.5 g)/(28.02 g/mol)

= 0.9458 mol

we have the Balanced chemical equation as:

3 H2 + N2 —> 2 NH3

3 mol of H2 reacts with 1 mol of N2

for 6.7808 mol of H2, 2.2603 mol of N2 is required

But we have 0.9458 mol of N2

so, N2 is limiting reagent

we will use N2 in further calculation

Molar mass of NH3 = 1*MM(N) + 3*MM(H)

= 1*14.01 + 3*1.008

= 17.034 g/mol

From balanced chemical reaction, we see that

when 1 mol of N2 reacts, 2 mol of NH3 is formed

mol of NH3 formed = (2/1)* moles of N2

= (2/1)*0.9458

= 1.892 mol

we have below equation to be used:

mass of NH3 = number of mol * molar mass

= 1.892*17.03

= 32.2 g

Answer: 32.2 g

2)

N2 is limiting reagent

3)

From balanced chemical reaction, we see that

when 1 mol of N2 reacts, 3 mol of H2 reacts

mol of H2 reacted = (3/1)* moles of N2

= (3/1)*0.9458

= 2.837 mol

mol of H2 remaining = mol initially present – mol reacted

mol of H2 remaining = 6.7808 – 2.8373

mol of H2 remaining = 3.9435 mol

Molar mass of H2 = 2.016 g/mol

we have below equation to be used:

mass of H2,

m = number of mol * molar mass

= 3.943 mol * 2.016 g/mol

= 7.95 g

Answer: 7.95 g of H2 remains