1 Answer

Sam Hokin · Answer 1 · 2020-07-23T01:43:15+0000

Answer : The mass of 1.50 mole of iron(III) sulfate is,

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 :

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

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

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

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

Therefore, the mass of 1.50 mole of iron(III) sulfate is,

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