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

`m_(NaNH_2)=30.42gNaNH_2`

`m_(H_2)=0.783gH_2`

Step-by-step 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*(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*(1molNH_3)/(17gNH_3) *(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*(2molNaNH_2)/(2molNa)*(39gNaNH_2)/(1molNaNH_2)=30.42gNaNH_2`

`m_(H_2)=0.783molNa*(1molH_2)/(2molNa)*(2gH_2)/(1molH_2)=0.783gH_2`

Best regards.