Answer:
The percentage yield is 80%
Step-by-step explanation:
First let us generate a balanced equation for the reaction. This is illustrated below:
MgCO3 —> MgO + CO2
Now, we proceed with our calculations as follows:
Molar Mass of MgCO3 = 24 + 12 + (16x3) = 24 + 12 + 48 = 84g/mol
Molar Mass of MgO = 24 + 16 = 40g/mol
From the equation,
84g of MgCO3 produced 40g of MgO.
Therefore, 50g of MgCO3 will produce = (50 x 40) /84 = 25g of MgO.
Now we can obtain the %yield as follows:
Experimental yield = 20g
Theoretical yield = 25g
%yield = Experimental yield /Theoretical yield x100
%yield = 20/25 x 100
%yield = 80%