Question

If you can answer both please do if not help with the top one please!

Answer 1

We are required to calculate the percentage yield of sodium carbonate.

We are given this reaction:

`2NaHCO_3\text{ }\rightarrow Na_2CO_3+H_2O+CO_2`

Given: mass of NaHCO3 = 1.357 g and molar mass =84.01g/mol

mass of Na2CO3 = 0.768g and molar mass = 105.99 g/mol

%Yield = (actual yield/theoretical yield) x 100

We do know the actual yield, which is 0.768 g, but we do not know the theoretical yield.

number of moles of NaHCO3 = 1.357 g/84.01 g/mol

=0.01615 mol

Take the number of moles and multiply by molar ratio and multiply by molar mass of Na2CO3 to the theoretical mass of Na2CO3.

mass of Na2CO3 = 0.01615 mol x (1/2) x 105.99 g/mol

= 0.856 g

%Yield = (0.768/0.856) x 100

%Yield = 89.7%

.