Question

Assuming that you start with 21.4 g of ammonia gas and 18.0 g of sodium metal and assuming that the reaction goes to completion, determine the mass (in grams) of each product.

Answer 1

Answer:

$m_{NaNH_2}=30.42gNaNH_2$

$m_{H_2}=0.783gH_2$

Explanation:

Hello,

In this case, the reaction between sodium and ammonia is:

$2Na+2NH_3\rightarrow 2NaNH_2+H_2$

Thus, as we know the initial masses of both sodium and ammonia, we should first identify the limiting reactant, for which we firstly compute the available moles of sodium:

$n_{Na}=18.0gNa*\frac{1molNa}{23.0gNa}=0.783molNa$

And the moles of sodium consumed by 21.4 g of ammonia (2:2 mole ratio):

$n_{Na}^{\ consumed}=21.4gNH_3*\frac{1molNH_3}{17gNH_3} *\frac{2molNa}{2molNH_3} =1.26molNa$

In such a way, since less moles of sodium are available than consumed by ammonia, we can say, sodium is the limiting reactant. Furthermore, the mass of both sodium amide (39 g/mol) and hydrogen gas (2 g/mol) that are produced turn out:

$m_{NaNH_2}=0.783molNa*\frac{2molNaNH_2}{2molNa}*\frac{39gNaNH_2}{1molNaNH_2}=30.42gNaNH_2$

$m_{H_2}=0.783molNa*\frac{1molH_2}{2molNa}*\frac{2gH_2}{1molH_2}=0.783gH_2$

Best regards.