The reaction of 52 g of NH 3 with excess oxygen yields 45 g of NO according to the balanced equation below. What is the percent yield? 4 NH3 + 5 O2 → 4 NO + 6 H2O a. 83% b. 100% c. 18% d. 25% e. 49%

To calculate the percent yield, you first need to calculate the theoretical yield of a reaction, which is the maximum amount of product that can be produced from the given quantity of reactants in a perfectly efficient reaction, and then compare it to the actual yield that you've obtained.

The balanced equation for the reaction is:

4 NH3 + 5 O2 → 4 NO + 6 H2O

First, calculate the moles of NH3 that reacted: The molar mass of NH3 is approximately 17 g/mol (N: 14 + H: 3). 52 g NH3 * (1 mol NH3 / 17 g NH3) = approximately 3.06 mol NH3.

Next, use the stoichiometry of the reaction to find out how many moles of NO would be formed from those moles of NH3 if the reaction had 100% yield. From the balanced equation, we can see the molar ratio between NH3 and NO is 1:1.

So the theoretical yield of NO in moles is the same as the amount of NH3 used: 3.06 mol NO.

Then convert moles of NO to grams to get the theoretical yield: The molar mass of NO is approximately 30 g/mol (N: 14 + O: 16). 3.06 mol NO * 30 g NO/mol = 91.8 g NO (theoretical yield).

The actual yield of NO is given as 45 g.

Now calculate percent yield using the formula:

Percent yield = (Actual yield / Theoretical yield) * 100%. Percent yield = (45 g / 91.8 g) * 100% = approximately 49%.

Therefore, the percent yield for this reaction is approximately 49%.

The best answer among the given choices is e. 49%.