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ololo11 [35]
4 years ago
14

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

Chemistry
1 answer:
lara31 [8.8K]4 years ago
8 0

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.

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