Question

What is the mass, in grams, of 1.50 mol of iron (III) sulfate?Express your answer using three significant figures.

Answer 1

Answer : The mass of 1.50 mole of iron(III) sulfate is, $5.99\times 10^2g$

Explanation : Given,

Moles of iron(III) sulfate = 1.50 mole

Molar mass of iron(III) sulfate = 399.88 g/mole

The formula of iron(III) sulfate is, $Fe_2(SO_4)_3$

Formula used :

$\text{Mass of }Fe_2(SO_4)_3=\text{Moles of }Fe_2(SO_4)_3\times \text{Molar mass of }Fe_2(SO_4)_3$

Now put all the given values in this formula, we get:

$\text{Mass of }Fe_2(SO_4)_3=1.50mole\times 399.88g/mole$

$\text{Mass of }Fe_2(SO_4)_3=599.82g=5.99\times 10^2g$

Therefore, the mass of 1.50 mole of iron(III) sulfate is, $5.99\times 10^2g$