Question

Nitric acid, a major industrial and laboratory acid, is produced commercially by the multi-step Ostwald process, which begins with the oxidation of ammonia:Step 1: NH3(g) + O2(g) ⟶ NO(g) + H2O(l)Step 2: NO(g) + O2(g) ⟶ NO2(g)Step 3: NO2(g) + H2O(l) ⟶ HNO3(l) + NO(g)Balance the equations. Assuming 100% yield in each step, what mass (in kg) of ammonia must be used to produce 7.137 kg of nitric acid.

Answer 1

Answer:

2.889 kg of ammonia must be used to produce 7.137 kg of nitric acid.

Explanation:

Step 1:

$4NH_3(g) + 5O_2(g)\rightarrow 4NO(g) + 6H_2O(l)$..[1]

Step 2:

$2NO(g) + O_2(g)\rightarrow 2NO_2(g)$..[2]

Step 3:

$3NO_2(g) + H_2O(l)\rightarrow 2HNO_3(l) + NO(g)$..[3]

3 × [1] + 6 × [2] + 4 × [3] = [4]

$12NH_3(g)+21O_2(g)\rightarrow 8HNO_3(g)+4NO(g)+14H_2O$..[4]

Mass of nitric acid = 7.137 kg = 7137 g( 1kg = 1000g)

Moles of nitric acid = $\frac{7137 g}{63 g/mol}=113.29 mol$

According reaction [4], 8 moles of nitric acid are obtained from 21 moles of ammonia . Then 113.29 moles of nitric acid will be obtained from:

$\frac{12}{8}\times 113.29 mol=169.935 mol$

Mass of 169.935 moles of ammonia;

169.935 mol × 17 g/mol = 2888.895 g

2888.895 g = 2888.895 0.001 kg =2.888895 kg ≈ 2.889 kg

2.889 kg of ammonia must be used to produce 7.137 kg of nitric acid.

Answer 2

In the multi-step Ostwald process, 2.895 kg of ammonia are required to synthesize 7.137 kg of nitric acid.

Let's consider the multi-step Ostwald process for the production of nitric acid.

• Step 1: 4 NH₃(g) + 5 O₂(g) ⟶ 4 NO(g) + 6 H₂O(l)

• Step 2: 2 NO(g) + O₂(g) ⟶ 2 NO₂(g)

• Step 3: 3 NO₂(g) + H₂O(l) ⟶ 2 HNO₃(l) + NO(g)

First, we will convert 7.137 kg of nitric acid to moles using its molar mass.

(7.137 × 10³ g) × (1 mol/63.01 g) = 113.3 mol

Next, we will calculate the moles of ammonia required to form 113.3 moles of nitric acid using the following conversion factors:

• From Step 3, the molar ratio of HNO₃ to NO₂ is 2:3.

• From Step 2, the molar ratio of NO₂ to NO is 2:2.

• From Step 1, the molar ratio of NO to NH₃ is 4:4.

$113.3 molHNO_3 \times \frac{3molNO_2}{2molHNO_3} \times \frac{2molNO}{2molNO_2} \times \frac{4molNH_3}{4molNO} = 170.0 mol NH_3$

Finally, we will convert 170.0 moles of NH₃ to mass using its molar mass.

170.0 mol × (17.03 × 10⁻³ kg/mol) = 2.895 kg

In the multi-step Ostwald process, 2.895 kg of ammonia are required to synthesize 7.137 kg of nitric acid.

Learn more about the Ostwald process here: https://brainly.com/question/15876986