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Hatshy [7]
3 years ago
7

What is the mass of silver chlorate (191.32 g/mol) that decomposes to release 0.466L of oxygen gas at STP? AgC1036) _AgCl). _026

) A) 1.33g B) 597 E) 7.968 C) 3.988 D) 2658
Chemistry
1 answer:
vesna_86 [32]3 years ago
3 0

Answer : The mass of silver chlorate will be 2.654 grams.

Explanation :

The balanced chemical reaction is,

2AgClO_3\rightarrow 2AgCl+3O_2

First we have to calculate the moles of oxygen gas at STP.

As, 22.4 L volume of oxygen gas present in 1 mole of oxygen gas

So, 0.466 L volume of oxygen gas present in \frac{0.466}{22.4}=0.0208 mole of oxygen gas

Now we have to calculate the moles of silver chlorate.

From the balanced chemical reaction, we conclude that

As, 3 moles of oxygen produced from 2 moles of silver chlorate

So, 0.0208 moles of oxygen produced from \frac{2}{3}\times 0.0208=0.01387 moles of silver chlorate

Now we have to calculate the mass of silver chlorate.

\text{Mass of }AgClO_3=\text{Moles of }AgClO_3\times \text{Molar mass of }AgClO_3

Molar mass of silver chlorate = 191.32 g/mole

\text{Mass of }AgClO_3=0.01387mole\times 191.32g/mole=2.654g

Therefore, the mass of silver chlorate will be 2.654 grams.

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