Under certain conditions the reaction below will proceed at 28.2% yield of NO. How many grams of NH3 must be reacted with excess oxygen to yield 125 g of NO?4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)

To solve this problem, we need to follow these steps:

1. First, we need to find the molar mass of NO and NH3. The molar mass of NO is approximately 30 g/mol (14 for Nitrogen and 16 for Oxygen) and the molar mass of NH3 is approximately 17 g/mol (14 for Nitrogen and 1 for Hydrogen x 3).

2. Next, we need to find out how many moles of NO are in 125 g. We do this by dividing the mass of NO by its molar mass. So, 125 g ÷ 30 g/mol = 4.17 mol of NO.

3. The reaction shows that 4 moles of NH3 produce 4 moles of NO. Therefore, the amount of NH3 needed is the same as the amount of NO produced, which is 4.17 mol.

4. However, the problem states that the reaction only proceeds at 28.2% yield. This means that we only get 28.2% of the maximum possible amount. So, to find out how much NH3 we actually need, we divide the amount of NO by the yield: 4.17 mol ÷ 0.282 = 14.79 mol of NH3.

5. Finally, to find out how many grams of NH3 this is, we multiply by the molar mass of NH3: 14.79 mol x 17 g/mol = 251.43 g.

So, we need approximately 251.43 g of NH3 to yield 125 g of NO under these conditions.