Question

How many moles are present in 560.0g of iron(III) sulfate?

A) 1.0 moles
B) 2.0 moles
C) 3.0 moles
D) 4.0 moles

Answer 1

To find the number of moles in 560.0g of iron(III) sulfate, we calculate the molar mass of the compound, which is 344.06 g/mol. By dividing the given mass by the molar mass, we find that there are approximately 1.626 moles of iron(III) sulfate.

Step-by-step explanation:

To determine the number of moles present in 560.0g of iron(III) sulfate, we need to use the molar mass of iron(III) sulfate and the given mass of the compound.

The molar mass of iron(III) sulfate (Fe2(SO4)3) is calculated as:

2(55.85 g/mol for Fe) + 3(32.07 g/mol for S) + 12(16.00 g/mol for O)

= 2(55.85 g/mol) + 3(32.07 g/mol) + 12(16.00 g/mol) = 55.85 g/mol + 96.21 g/mol + 192.00 g/mol = 344.06 g/mol

Now we can use the molar mass to convert grams to moles:

Number of moles = Mass (g) / Molar mass (g/mol)

Number of moles = 560.0 g / 344.06 g/mol = 1.626 moles

Therefore, the answer is approximately 1.626 moles (rounded to three significant figures).