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200 g of ammonium sulfate reacts with 190 g of calcium carbonate. How many grams of calcium sulfate are produced?

A. 204 g
B. 190 g
C. 200 g
D. 150 g

Answer :

Let's start by identifying the compounds.

Ammonium sulfate has the ammonium cation (NH₄⁺) and the sulfate anion (SO₄²⁻). Thus the copound is: (NH₄)₂SO₄.

Calcium carbonate has calcium cation (Ca²⁺) and the carbonate anion (CO₃²⁻). So, the compound is: CaCO₃.

The reaction will be the exchange of the cations and anions:

[tex](NH_4)_2SO_4+CaCO_3\to CaSO_4+(NH_4)_2CO_3[/tex]

To form the calccium sulfate we want, CaSO₄.

The reaction is 1:1:1:1.

Let's now calculate how many moles we have of each reactant.

First, the molar weight of them:

Ammonium sulfate (NH₄)₂SO₄:

[tex]M_{\mleft(NH_4\mright)_2SO_4}=2M_N+8M_H+1M_S+4M_O\approx132.14g/mol[/tex]

Calcium carbonate CaCO₃:

[tex]M_{CaCO_{3}}=1M_{Ca}+1M_C+3M_O\approx100.09g/mol[/tex]

Now, the number of moles of each:

[tex]n_{\mleft(NH_4\mright)_2SO_4}=\frac{m_{\mleft(NH_4\mright)_2SO_4}}{M_{\mleft(NH_{4}\mright)_{2}SO_{4}}}=\frac{200g}{132.14g/mol}=1.5135\ldots g/mol[/tex]

[tex]n_{CaCO_3}=\frac{m_{CaCO_3}}{M_{CaCO_{3}}}=\frac{190g}{100.09g/mol}=1.8982\ldots g/mol[/tex]

Since the reaction is 1:1 for the reactants and we have more (NH₄)₂SO₄ than CaCO₃, the (NH₄)₂SO₄ is the limitant. So, at most, approximately 1.

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