Question

What is the mass of aluminum oxide (101.96 g/mol) produced from 1.74 g of manganese(IV) oxide (86.94 g/mol)?

Answer 1

Answer on Question #71061 – Chemistry – Other

Task:

What is the mass of aluminum oxide (101.96 g/mol) produced from 1.74 g of manganese (IV) oxide (86.94 g/mol)?

Solution:

The reaction equation:

$$3 M n O _ {2} + 4 A l = 2 A l _ {2} O _ {3} + 3 M n$$

3 mol MnO₂ produce 2 mol Al₂O₃.

By the reaction equation: $\frac{n(MnO_2)}{3} = \frac{n(Al_2O_3)}{2}$ .

$$n = \frac {m}{M}.$$

Then,

$$\frac {m (M n O _ {2})}{3 * M (M n O _ {2})} = \frac {m (A l _ {2} O _ {3})}{2 * M (A l _ {2} O _ {3})};$$ $$m (A l _ {2} O _ {3}) = \frac {2 * M (A l _ {2} O _ {3}) * m (M n O _ {2})}{3 * M (M n O _ {2})};$$ $$m (A l _ {2} O _ {3}) = \frac {2 * 1 0 1 . 9 6 \frac {g}{m o l} * 1 . 7 4 g}{3 * 8 6 . 9 4 \frac {g}{m o l}} = 1. 3 6 g.$$

Answer: $1.36\mathrm{gAl_2O_3}$ produced.

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