When nitrogen dioxide gas dissolves in water, an aqueous solution containing dissolved nitric acid and nitrogen monoxide forms. In terms of mass balance, 138.03 g of NO_2 completely reacts with 18.02 g of H_2O to form 126.04 g of HNO_3 and 30.01 g of NO. 3NO_2 (g) + H_2O (l) rightarrow 2HNO_3 (aq) + NO (aq) 1. When 359 g of NO_2 is used, how many grams of H_2O are consumed? 2. How many grams of HNO_3 are produced? 3. How many grams of NO are produced?

Question

Chemistry

Matilda Wilkerson

7 June, 10:20

When nitrogen dioxide gas dissolves in water, an aqueous solution containing dissolved nitric acid and nitrogen monoxide forms. In terms of mass balance, 138.03 g of NO_2 completely reacts with 18.02 g of H_2O to form 126.04 g of HNO_3 and 30.01 g of NO. 3NO_2 (g) + H_2O (l) rightarrow 2HNO_3 (aq) + NO (aq) 1. When 359 g of NO_2 is used, how many grams of H_2O are consumed? 2. How many grams of HNO_3 are produced? 3. How many grams of NO are produced?

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Answers (1)

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Laylah Foley · Answer 1

1. Based on the given question, 138.03 grams of NO₂ is reacting completely with 18.02 grams of H2O. However, in case when 359 grams of NO₂ is used then the grams of water consumed in the reaction will be,

= 359 * 18.02 / 138.03 = 46.87 grams of water.

2. As mentioned in the given case 138.03 grams of NO₂ generates 126.04 grams of HNO₃. Therefore, 359 grams of NO₂ will produce,

= 359 * 126.04 / 138.03

= 327.81 grams of HNO₃.

3. Based on the given question, 138.04 grams of NO₂ is generating 30.01 grams of NO. Therefore, 359 grams of NO₂ will generate,

= 359 * 30.01 / 138.04

= 78.04 grams of NO.