The reaction between 34.0 g of NH3 and excess oxygen gas produces 45.9 g of NO gas and some water. Determine the percent yield.

See your post above.

77 percent

To determine the percent yield of a reaction, we need to compare the actual yield to the theoretical yield.

First, let's calculate the theoretical yield of NO gas. We can do this by using the balanced equation for the reaction:

4NH3 + 5O2 -> 4NO + 6H2O

From the equation, we can see that the molar ratio between NH3 and NO is 4:4, which means that 1 mole of NH3 will produce 1 mole of NO.

1 mole of NH3 has a molar mass of 17.03 g. So, 34.0 g of NH3 is equal to 34.0 g / 17.03 g/mol = 1.9963 mol of NH3.

Since the molar ratio between NH3 and NO is 1:1, the theoretical yield of NO gas is also 1.9963 mol.

The molar mass of NO is 30.01 g/mol. So, the theoretical yield of NO gas in grams is 1.9963 mol x 30.01 g/mol = 59.91 g.

Now, let's calculate the percent yield by comparing the actual yield to the theoretical yield:

Percent Yield = (Actual Yield / Theoretical Yield) x 100

Given that the actual yield of NO gas is 45.9 g, we can substitute these values into the equation:

Percent Yield = (45.9 g / 59.91 g) x 100 = 76.7%

Therefore, the percent yield of the reaction is approximately 76.7%.