Question

What is the mass, in grams, of 0.0490 mol of iron(III) phosphate

Answer 1

Answer:

m = 7.39 g.

Explanation:

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In this case, since the molar mass of iron (III) phosphate is 150.82 g/mol based on its molecular formula (FePO₄), we can compute the mass in grams of 0.0490 moles of this compound by setting up the following dimensional analysis:

$m=0.0490mol*\frac{150.82g}{1mol} \\\\m=7.39 g$

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