Examine the statement and balanced equation.

Pure elemental iron undergoes a single replacement reaction with sulfuric acid [tex]\((H_2SO_4)\)[/tex] to produce iron(III) sulfate [tex]\((Fe_2(SO_4)_3)\)[/tex] and hydrogen gas.

[tex]\[2Fe + 3H_2SO_4 \rightarrow Fe_2(SO_4)_3 + 3H_2\][/tex]

[tex]\((M_{Fe} = 55.8 \, \text{g/mol}, \, M_{H_2SO_4} = 98.1 \, \text{g/mol}, \, M_{Fe_2(SO_4)_3} = 399.9 \, \text{g/mol}, \, M_{H_2} = 2.0 \, \text{g/mol})\)[/tex]

If [tex]\(3.21 \times 10^{12} \, \text{mol}\)[/tex] of iron(III) sulfate are formed, how many grams of it will be produced?

A. [tex]\(1.28 \times 10^{15} \, \text{g} \, Fe_2(SO_4)_3\)[/tex]

B. [tex]\(3.24 \times 10^{14} \, \text{g} \, Fe_2(SO_4)_3\)[/tex]

C. [tex]\(8.02 \times 10^9 \, \text{g} \, Fe_2(SO_4)_3\)[/tex]

D. [tex]\(3.09 \times 10^{10} \, \text{g} \, Fe_2(SO_4)_3\)[/tex]

To find the mass of iron(III) sulfate [tex]\((Fe_2(SO_4)_3)\)[/tex] produced from the given moles, we need to follow these simple steps:

1. Identify the number of moles of iron(III) sulfate given:
- We have [tex]\(3.21 \times 10^{12}\)[/tex] moles of iron(III) sulfate.

2. Determine the molar mass of iron(III) sulfate:
- The molar mass of [tex]\(Fe_2(SO_4)_3\)[/tex] is given as 399.9 g/mol.

3. Calculate the mass:
- To calculate the mass of iron(III) sulfate produced, multiply the number of moles by the molar mass:
[tex]\[
\text{mass of } Fe_2(SO_4)_3 = (\text{number of moles}) \times (\text{molar mass})
\][/tex]
- Plugging in the values:
[tex]\[
\text{mass of } Fe_2(SO_4)_3 = 3.21 \times 10^{12} \, \text{mol} \times 399.9 \, \text{g/mol}
\][/tex]

4. Result:
- The mass of [tex]\(Fe_2(SO_4)_3\)[/tex] produced is [tex]\(1.28 \times 10^{15}\)[/tex] grams.

Therefore, the correct answer is [tex]\(1.28 \times 10^{15}\)[/tex] g [tex]\(Fe_2(SO_4)_3\)[/tex].