Magnesium carbonate decomposes on heating to form magnesium oxide and carbon dioxide. Suppose that 50.0 g of magnesium carbonate was heated, and that 20.0 g of magnesium oxide was subsequently collected. What was the percent yield of this reaction?

The percentage yield is 80%

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%

14 days is the half-life of phosphorus-32

6.4 cm (max) - 25 cm (min)