Question

What mass of NH3 (in grams) must be used to produce 5.65 tons of HNO3 by the Ostwald process, assuming an 80.0 percent yield in each step (1 ton = 2000 lb; 1 lb = 453.6 g)? Enter your answer in scientific notation.

Answer 1

Answer:

The answer to the question is

The mass of NH3 (in grams) that must be used to produce 5.65 tons of HNO3 by the Ostwald process, assuming an 80.0 percent yield in each step in scientific notation is 3.46 × 10⁶ grams.

Explanation:

The reaction for the Ostwald process is given by

4NH₃ + 5O₂  → 4NO + 6H₂O

2NO + O₂  ↔ 2NO₂

3NO₂ (g) + H₂O (l) → 2HNO₃ + NO

From the reactions of the Ostwald process, 4 moles of NH₃ are required to produce 2 moles of HNO₃

Therefore One mole of HNO₃ requires two moles of NH₃  for production at 100 % theoretical yield

Molar mass of NH₃  =  17.031 g/mol

Molar mass of HNO₃  = 63.01 g/mol

5.65 tons = 5.65 × 2000 lb/ton × 453.6 g/lb = 5125680 g

Number of moles of HNO₃ in 5125680 g sample of HNO₃

= (5125680 g)÷(63.01 g/mol) = 81347.09 moles of HNO₃

However the yield = 80.0%  that is $\frac{Actual Yield}{Theoretical Yield} 100$ = 80 %

Therefore 80 % Theoretical yield = 81347.09 moles of HNO₃

Theoretical yield = 81347.09 moles ÷ 0.8  = 101683.86 moles

Then 101683.86 moles of HNO₃ will require 2 × 101683.86 moles or 203367.72 moles of NH₃

Mass of 203367.72 moles of NH₃  is given by

mass = number of moles × molar mass = 203367.72 moles ×  17.031 g/mol = 3463555.63 grams

Mass of NH₃ required = 3463555.63 grams = 3.46 × 10⁶ grams  in scientific notation.