Question

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

Answer 1

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
`(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
`(2)/(3)* 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* \text{Molar mass of }AgClO_3`

Molar mass of silver chlorate = 191.32 g/mole

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

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