Consider the following reaction: 2N205 --> 4NO2(g) + 02(g)
When 40 g of N2O5 decomposes, 4.5 g of 02 is formed. What is the percent
yield?
To calculate the percent yield, we need to compare the actual yield to the theoretical yield.
First, let's calculate the theoretical yield of O2:
From the balanced chemical equation, we can see that for every 2 moles of N2O5, we would expect to produce 1 mole of O2.
The molar mass of N2O5 is 108.02 g/mol, so 40 g is equal to 40/108.02 = 0.370 mol of N2O5.
Therefore, the theoretical yield of O2 would be (0.370 mol N2O5) × (1 mol O2 / 2 mol N2O5) × (32.00 g/mol O2) = 5.92 g O2.
The percent yield can be calculated using the following formula:
Percent Yield = (Actual Yield / Theoretical Yield) × 100
In this case, the actual yield of O2 is given as 4.5 g.
Plugging in the values, we get:
Percent Yield = (4.5 g / 5.92 g) × 100 = 76.01%
Therefore, the percent yield of O2 in this reaction is approximately 76.01%.