Question

What mass of iron(II) sulfate would be needed to provide 28 grams of iron?

Answer 1

Answer:

76 grams

Explanation:

$mol = mass/mr$

$mol \: fe = 28 \div 56 = 0.5$

$mass = mol \times mr$

$mol \times mr = 0.5 \times 152$

$0.5 \times 152g = 76g$