Question

The chemical equation shows how ammonia reacts with sulfuric acid to produce ammonium sulfate. 2NH3(aq) + H2SO4(aq) (NH4)2SO4(aq) How many grams of ammonium sulfate can be produced if 60.0 mol of sulfuric acid react with an excess of ammonia? 1,020 g 3,970 g 5,890 g 7,930 g

Answer 1

Ans: D) 7930 g

Given:

Moles of H2SO4 = 60.0 mol

NH3 in excess

To determine:

The amount in grams of (NH4)2SO4 produced

Explanation:

The chemical reaction is as follows:

2NH3 + H2SO4 → (NH4)2SO4

Since, NH3 is in excess, H2SO4 will be the limiting reagent and will influence the amount of product formed

Based on the reaction stoichiometry:

1 mole of H2SO4 forms 1 mole of (NH4)2SO4

Therefore, 60.0 mol of H2SO4 will produce 60.0 mol of (NH4)2SO4

Now:

$moles = \frac{mass}{molar mass}$

molar mass (NH4)2SO4 = 132.14 g/mol

$mass of (NH4)2SO4 = moles * molar mass\\\\= 60.0 mol * 132.14 g/mol = 7928.4 g$

Answer 2

D. 7,930 g ................................ believe me