How many moles of each substance is produced when 7 moles of ammonium carbonate decompose? (NH4)2CO3(s) ⟶ NH3(g) + CO2(g) + H2O(g)

1. 14 moles of NH3 were produced.

2. 7 moles of CO2 were produced.

3. 7 moles of H2O were produced.

Explanation:

We'll begin by writing a balanced equation for the decomposition of ammonium carbonate, (NH4)2CO3.

This is illustrated below:

(NH4)2CO3(s) —> NH3(g) + CO2(g) + H2O(g)

There are 2 atoms of N on the left side and 1 atom on the right side. It can be balance by putting 2 in front of NH3 as shown below:

(NH4)2CO3(s) —> 2NH3(g) + CO2(g) + H2O(g)

Now the equation is balanced.

From the balanced equation above,

1 mole of (NH4)2CO3 decompose to produce 2 moles of NH3, 1 mole of CO2 and 1 mole of H2O.

1. Determination of the number of mole of NH3 produced.

From the balanced equation above,

1 mole of (NH4)2CO3 decompose to produce 2 moles of NH3.

Therefore, 7 moles of (NH4)2CO3 will decompose to produce = 7 × 2 = 14 moles of NH3.

Therefore, 14 moles of NH3 were produced.

2. Determination of the number of mole CO2 produced.

From the balanced equation above,

1 mole of (NH4)2CO3 decompose to produce 1 mole of CO2.

Therefore, 7 moles of (NH4)2CO3 will also decompose to produce 7 moles of CO2.

Therefore, 7 moles of CO2 were produced.

3. Determination of the number of mole of H2O produced.

From the balanced equation above,

1 mole of (NH4)2CO3 decompose to produce 1 mole of H2O.

Therefore, 7 moles of (NH4)2CO3 will also decompose to produce 7 moles of H2O.

Therefore, 7 moles of H2O were produced.

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