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

Respuesta :

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

Explanation :

The balanced chemical reaction is,

[tex]2AgClO_3\rightarrow 2AgCl+3O_2[/tex]

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 [tex]\frac{0.466}{22.4}=0.0208[/tex] 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 [tex]\frac{2}{3}\times 0.0208=0.01387[/tex] moles of silver chlorate

Now we have to calculate the mass of silver chlorate.

[tex]\text{Mass of }AgClO_3=\text{Moles of }AgClO_3\times \text{Molar mass of }AgClO_3[/tex]

Molar mass of silver chlorate = 191.32 g/mole

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

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