I need some help on this question, i dont quite understand

Question

If 1.70 g of Nitrogen react with 1.09 g of hydrogen to produce ammonia, what is the limiting reactant and what mass of ammonia is produced?

a. Hydrogen is the limiting reactant and 0.22g of ammonia are produced.

b. Nitrogen is the limiting reactant and 2.07g of ammonia are produced.

c. Nitrogen is the limiting reactant and 4.15g of ammonia are produced.

Answer 1

Hi Rosemary L.,

When you need to find limiting reactants, you need to be comparing the reactants in terms of moles of each. That's because they react stoichiometrically by atoms / molecules / moles (all are correct units, which one you use depends on the particular question.)

In this case, you need to 1) convert each given into moles of THAT material. So:

H = 1.09g / (1 g/mole) = 1.09 mole

N = 1.70g / (14 g/mole) =~ 0.1214 mole

2) then figure the product, first as moles, then as mass: since 3 H are needed to react with 1 N, that would require only 3*0.1214 = 0.3642 mole H to completely use up the N. So H is in excess. Then you can make 0.1214 mole NH3 or about (*17 g/mole) =~ 2.06 g NH3 .

So study the strategy above, and remember tro always be ready to convert mass->moles and moles->mass!

-- Cheers, --Mr. d.

Answer 2

First write the UNBLANCED chemical reaction equation:

Since hydrogen and nitrogen and diatomic gasses it would be:

N₂ + H₂ → NH₃

Now balance the equation:

N₂ + 3H₂ → 2NH₃

Now convert the grams of nitrogen and hydrogen into moles by first calculating the molar mass (looking up each element in the periodic table)::

N₂ = 14.007 x 2 = 28.014 g/mol

H₂ = 1.008 x 2 = 2.016 g/mol

(1.70 g N₂)/(28.014 g/mol) = 0.06068 moles

(1.09 g of H₂)/(2.016 g/mol) = 0.5407 moles

The balanced chemical reaction equation says we need 1 mole of N₂ for every 3 moles of H₂ so obviously there is more H₂ than necessary to react with the 0.06068 moles of N₂ so N₂ (nitrogen) is the limiting reactant. The balanced chemical reaction equation also tells us that 2 moles of ammonia (NH₃) are produced from every 1 mole of N₂. So if we start with 0.06068 moles of N₂, we will end with 0.1214 moles of NH₃.

To determine the mass of 0.1214 moles of NH₃ calculate the molar mass:

N: 14.007 x 1 = 14.007

H: 1.008 x 3 = 3.024

NH₃ = 17.031 g/mol

The multiply 0.1214 moles of NH₃ by 17.031 g/mol to get 2.07 g of NH₃