Question

Determine the limiting reactant (LR) and the mass (in g) of nitrogen that can be formed from 50.0 g N 2O 4 and 45.0 g N 2H 4. Some possibly useful molar masses are as follows: N 2O 4 = 92.02 g/mol, N 2H 4 = 32.05 g/mol.

N 2O 4( l) + 2 N 2H 4( l) → 3 N 2( g) + 4 H 2O( g)

		LR = N2O4, 45.7 g N2 formed
		LR = N2O4, 105 g N2 formed
		LR = N2H4, 13.3 g N2 formed
		LR = N2H4, 59.0 g N2 formed
		No LR, 45.0 g N2 formed

Answer 1

Firslty calculate number of moles of N2O4 and N2H4

No. Of moles of n2o4 = given mass of n2o4/ molar mass of N2O4

= 50 g / 92.02 g/mol = 0.543mole

No. Of moles of N2h4 = given mass of n2h4/ molar mass of n2h4= 45g/32.05g/mol = 1.404 mole

According to the given reaction

2 mole of n2h4 reacts with 1 mole of n2o4

1 moles of n2h4 reacts with 1/2 mole of n2o4

1.404 moles of n2h4 reacts with 1/2* 1.404 = 0.702mole of n2o4

But we have only 0.543 mole of n2O4

It means n2o4 is a limiting reagent

And the formation of product depends on the amount of limiting reagent

According to the balanced given reaction

1 mole of n2o4 forms 3 mole of N2

0.543 mole of n2o4 form 3*0.543 mole of n2= 1.629 mole of N2

To convert 1.629 mole of N2 into g

Multiply it with molar mass of N2 (28g/mole)

1.629 mole * 28g/mole =45.612g

So the option 1 is correct

Hope it will be helpful to u

Thanku