Question

Consider the following balanced reaction. How many grams of water are required to form 75.9 g of HNO3? Assume that there is excess NO2 present. The molar masses are as follows: H2O = 18.02 g/mol, HNO3 = 63.02 g/mol. 3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g)

Answer 1

Answer:

The answer is "10.84 g".

Explanation:

The formula for calculating the number for moles:

$\text{Number of moles }= \frac{\ Mass}{ \ molar \ mass }$

In the given acid nitric:

Owing to the nitric acid mass = $75.9 g$

Nitric acid molar weight$= 63\ \frac{g}{mol}$

If they put values above the formula, they receive:

$\text{moles in nitric acid} = \frac{75.9}{63}$

                               $=1.204 \ mol$

In the given chemical equation:

$3 NO_2 \ (g) + H_2O \ (l) \longroghtarrow 2 HNO_3 \ (aq) + NO\ (g)$

In this reaction, 2 mols of nitric acid are produced by 1 mole of water.

So, 1.204 moles of nitric acid will be produced:

$= \frac{1}{2} \times 1.204$

$=0.602\ \ \text{mol of water}.$

We are now using Equation 1 in determining the quantity of water:

Water moles $= 0.602\ mol$

Water weight molar $= 18.02 \ \frac{g}{mol}$

$\to 0.602 = \frac{\text{mass of mols}}{ 18.02}\\\\\to \text{mass of mols} = 0.602 \times 18.02$

                         $=10.84 \ g$