Nitrogen dioxide reacts with water to form nitric acid and nitrogen monoxide according to the equation:3NO2(g)+H2O(l)→2HNO3(l)+NO(g) Suppose that 4.5 mol NO2NO2 and 0.40 mol H2OH2O combine and react completely. Which reactant is in excess?

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orchidmammoth756Lv1

1 Jun 2020

Nitrogen dioxide reacts with water to form nitric acid and nitrogen monoxide according to the equation:
$3 N O_{2} (g) + H_{2} O (l) \to 2 H N O_{3} (l) + N O (g)$

$3 N O_{2} (g) + H_{2} O (l) \to 2 H N O_{3} (l) + N O (g)$

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Jamar FerryLv2

1 Jun 2020

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Related questions

Some of the acid in acid rain is produced by the following reaction:

3NO2(g) + H2O(l) ----> 2HNO3(aq) + N (g)

(a) How many moles of water (H2O) would be converted to acid rain from 150 g of NO2 (46.0g/mol)?

(b) If 150.0g of NO2 is mixed with 15.0 grams of H2O, how many moles of which reactant remains?

The Ostwald process for the commercial production of nitric acid from ammonia and oxygen involves the following steps:

4NH3(g)+5O2(g)------->4NO(g)+6H2O(g) -908 kj/mol (delta H)

2NO(g)+O2(g)----------->2NO2(g) -112 kj/mol

3NO2(g)+H2O(l)-------->2HNO3(aq)+NO(g) -140 kj/mol

Write the overall equation for the production of nitric acid by the Ostwald process by combining the preceding equations (Water is also a product).

So i know the answer is this...

"Multiply the equations so that the coefficients of the N-products become the same as the N-educts of the following line.
4NH3(g)+5O2(g)------->4NO(g)+6H2O(g) -908 kj/mol (delta H) (x3)
2NO(g)+O2(g)----------->2NO2(g) -112 kj/mol (x 6)
3NO2(g)+H2O(l)-------->2HNO3(aq)+NO(g) -140 kj/mol (x 4)

12NH3(g)+15O2(g)------->12NO(g)+18H2O(...
12NO(g)+6O2(g)----------->12NO2(g)
12NO2(g)+4H2O(l)-------->8HNO3(aq)+4NO.."

What i want to know is why do we multiply the equations like that? Am i missing a concept? Because why can we just not add the eqautions together?