Question

Calculate the moles of O atoms in 0.658 g of Mg(NO3)2.

Answer 1

Answer:

There are 0.0267 moles of oxygen in 0.658 grams of magnesium nitrate.

Explanation:

Mass of magnesium nitrate = m = 0.658 g

Molar mass of magnesium nitrate = M = 148 g/mol

Moles of magnesium nitrate = n

$n=\frac{m}{M}=\frac{0.658 g}{148 g/mol}=0.004446 mol$

1 mole of magnesium nitrate has 6 moles of oxygen atoms. Then 0.004446 moles magnesium nitrate will have :

$6\times 0.004446 mol=0.026676 mol\approx 0.0267$

There are 0.0267 moles of oxygen in 0.658 grams of magnesium nitrate.