# 2NH_3(aq) + H_2SO_4(aq) -> (NH_4)_2SO_4(aq). How many grams of ammonium sulfate can be produced if 60.0 mol of sulfuric acid react with an excess of ammonia?

Jun 10, 2017

Assuming that there is an excess in ammonia, there are 7962 grams of of ammonium sulfate that can be produced for every 60 moles of sulfuric acid that reacts. More on that below.

#### Explanation:

What you are trying to find is called the theoretical yield, or how much of a particular product can we expect to form from a given amount of a reactant, assuming that the given reactant reacts completely.

Of course, this does not happen in reality where substances don't fully react therefore this is called the theoretical yield.

When you calculate the theoretical yield, you are actually asking:
How much of the product can I produce using all of the given reactant?
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In this case, the question will be:
How much ammonium sulfate can I form if I use all the 60 moles of sulfuric acid?
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In order to answer this question, there are several steps involved:

## 1. Identify the givens and what you need to find

• You are given 60 moles of sulfuric acid
• An excess amount of Ammonia

## 2. Use the balanced equation's ratios (if the equation isn't balanced, balance it) and apply the ratios to the problem

The balanced equation is like a recipe.

For example, in order to make one sandwich, there must be two slices of bread and one slice of ham. In other words:

**2 Slices of Bread + 1 Slice of Ham ==> 1 sandwich

** To make a dozen sandwiches, simply multiply both sides by 12 which will change the ratios proportionally

24 Slices of Bread + 12 Slices of Ham ==> 12 sandwiches

A mole is like a dozen, it equals Avogadro's Number which is approximately 6.022 × 10^23, so you would need to multiply both sides by this number in order to express the number in moles. This large number is simply expressed as "moles."

2 Moles of Slices of Bread + 1 Mole of Slices of Ham ==> 1 Mole of Sandwiches

In chemical equations, the same applies. The number of substances in a chemical equation are assumed to be in moles. In this problem, we can see: 2NH3 + H2SO4 → (NH4)2SO4

That 2 moles of NH3 combined with 1 mole of H2SO4 produces one mole of (NH4)2SO4 So for every mole of H2SO4 that reacts, we produce one mole of (NH4)2SO4

So it is a 1:1 (one-to-one ratio of NH3 and (NH4)2SO4) So we can similarly conclude that it takes 60 moles of NH3 to produce 60 moles of (NH4)2SO4

Therefore, we produce: 60 moles of (NH4)2SO4
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## 3. Calculate the molar mass of the product (NH4)2SO4

Molar mass, or grams per mole of each element is found in the periodic table. In order to calculate the molar mass of a substance that includes more than one element, we simply add the molar mass of each element that makes-up the substance:

$\text{2(N + 4H) + S + 4(O) = 2N + 8H + S + 4O = 2(14.01) + 8(1.01) + 32.07 + 4(16.00)}$

${\text{= 132.17 g/mol of" ("NH"_4)_2"SO}}_{4}$

(This will be your conversion factor of grams to moles)

## 4. Finally, convert the moles of product to grams of product

We have sixty moles of ("NH"_4)_2"SO"_4 to convert into grams:

And we need to find x grams of ("NH"_4)_2"SO"_4. The grams of ("NH"_4)_2"SO"_4 in the numerator of the first fraction cancels out with the denominator of the second fraction which leaves us with

$60 \times {\text{132.7 g of (NH"_4)_2"SO}}_{4}$

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