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?

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.  

the answer is b. dilution