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If 146.5 g of chromium(lll) chlorate decomposes, what mass of oxygen can be produced?

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Answer:

The Equation for this reaction is as follow,

2 Cr(ClO₃)₃ → 2 CrCl₃ + 9 O₂

According to eq,

604.68 g (2mole) of Cr(ClO₃)₃ produces = 288 g (9 moles) of O₂

So,

146.5 g of Cr(ClO₃)₃ will produce = X g of O₂

Solving for X,

X = (146.5 g × 288 g) ÷ 604.68 g

X = 69.77 g of O₂

User Ian MacFarlane
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