Consider the following 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. no2(g) + h2o(l) → hno3(aq) + no(g)

The mass of water required will be 10.848 g.

Explanation:

To calculate the number of moles, we use the equation:

   ......(1)

Given mass of nitric acid = 75.9 g

Molar mass of nitric acid = 63 g/mol

Putting values in above equation, we get:

For the given chemical equation:

By Stoichiometry of the reaction:

2 moles of nitric acid is produced by 1 mole of water.

So, 1.204 moles of nitric acid will be produced by = of water.

Now, calculating the amount of water, we use equation 1:

Moles of water = 0.602 mol

Molar mass of water = 18.02 g/mol

Putting values in equation 1, we get:

Hence, the mass of water required will be 10.848 g.