Question

.13 g H2 is allowed to react with 9.56 g N2, producing 2.92 g NH3. Part A What is the theoretical yield for this reaction under the given conditions?

Answer 1

The chemical reaction between N2 and H2 is as follows:

3 H2 + N2 >> 2 NH3

Now clacualte the number of moles of H2 and N2:

Moles H2 =amount in g/ molar mass

= 1.13 g / 2.016 g/mol = 0.56
Moles N2 = amount in g/ molar mass

=9.56 g / 28.0134 g/mol = 0.34

Moles H2 required to reacts N2 = 3 x 0.34 = 1.02
here 0.56 moles of H2 so it is the limiting reactant

Now calculate the number of moles of NH3:
0.56 moles of H2 * 2 mole NH3 /3 moles of H2

moles NH3 = 0.37 Moles NH3

Now calculate the mass of NH3:

Mass NH3 = 0.37 mole NH3 x 17.0307 g/mol =6.3 g

% yield =observed yield / theoretical yield *100

% yield = 2.92 x100 / 6.3 g

= 46.3%