Question

Помогите пожалуйста Избытком концентрированной азотной кислоты подействовали на 5,6 литров аммиака при этом образовалось 16 г нитрата аммония.Какова масовая доля выхода NH4NO3 в % от теоретически возможного выхода?

Answer 1

NH3+HNO3=NH4NO3
M(NH4NO3)=80г/моль
n(NH3)=5,6/22,4=0,25моль
n(NH4NO3)=0,25моль
m(NH4NO3)=0,25*80=20г
w=16*100/20=80%

Answer 2

Calculating Theoretical Yield

To calculate the theoretical yield of NH4NO3, we first need to balance the chemical equation: 5.6 L NH3 + excess HNO3 → 16 g NH4NO3 The molar amounts are: NH3: 5.6 L * 1 mol/17 g/L = 0.328 mol HNO3: excess NH4NO3: 16 g / 80 g/mol = 0.2 mol

The balanced equation is: 0.328 mol NH3 + 0.328 mol HNO3 → 0.2 mol NH4NO3

The theoretical yield of NH4NO3 is 0.2 mol or 16 g.

Calculating Actual Yield

The question states that 16 g of NH4NO3 was obtained.

Calculating Percent Yield

Percent yield is calculated as: Actual Yield / Theoretical Yield x 100% 16 g / 16 g x 100% = 100%

Therefore, the mass percent yield of NH4NO3 is 100% of the theoretical yield.