Question

In the reaction below, 7.0 mol of NO and 5.0 mol of 0, are reacted together. The reaction generates 3.7 mol of NO, What is the percent yield for the reaction? 2 NO (g) + O2 (g) → 2 NO2 (g)

Answer 1

The given amount of compounds that we have is:-

7.0 mol of NO

5.0 mol of O2

3.7 mol of NO2 as product, which is also the actual yield

Balanced chemical equation:-

2NO(g) + O2(g) ----> 2NO2(g)

It is evident from the balanced chemical equation that 2 moles of NO reacts with 1 mole of O2 to produce 2 moles of NO2.

For 7.0 mol of NO 3.5 mol of O2 is required, which means that NO is limiting reagent here in the above chemical reaction.

Hence, 7.0 mol of NO will produce 7.0 mol of NO2.

Therefore,

Theoretical yield = 7.0 mol

Actual yield = 3.7 mol -----(given amount of product)

We will find the % yield using the formula,

PercentYield = actualyield theoreticalvidd X 100

37 X 100

= 0.5286 X 100

Percent Yield= 52.86%

The Percent yield for the reaction is 52.86%.