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What is the mass of 2.56 × 10–4 moles of Fe2O3? : 6.23 × 105 159.6 g 1.60 × 0–6 g 4.09 × 10–2 g

User Gohmz
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1 Answer

5 votes

Answer:

4.09 × 10⁻² g

Solution:

As we know that,

1 mol of Fe₂O₃ is equal to = 159.69 g

So,

2.56 × 10⁻⁴ mol will be equal to = X g

Solving for X,

X = (159.69 g × 2.56 × 10⁻⁴ mol) ÷ 1 mol

X = 0.04088 g

Or,

X = 0.049 g

Or,

X = 4.09 × 10⁻² g

User Andrei Sosnin
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