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3 years ago

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Calculate the volume of oxygen at NTP obtained by decomposing 12.26g of KCLO3(at wt. K=39.1, Cl=35.5 and O = 16)

1 answer:

Anuta_ua [19.1K]3 years ago

8 0

Answer: 3.61 L

Explanation:

To calculate the moles, we use the equation:

$moles=\frac{\text {given mass}}{\text {Molar mass}}$

$moles=\frac{12.26g}{122.6g/mol}=0.1moles$

$2KClO_3\rightarrow 2KCl+3O_2$

2 moles of $KClO_3$ produce = 3 moles of $O_2$

0.1 moles of $KClO_3$ produce = $\frac{3}{2}\times 0.1=0.15$ moles of $O_2$

According to the ideal gas equation:'

$PV=nRT$

P = Pressure of the gas = 1 atm (NTP)

V= Volume of the gas = ?

T= Temperature of the gas = 20°C = (20+273) K = 293 K (NTP)

R= Value of gas constant in in kilopascals = 0.0821 Latm/K mol

$1\times V=0.15\times 0.0821\times 293$

$V=3.61L$

Thus volume of oxygen at NTP obtained by decomposing 12.26 g of $KClO_3$ is 3.61 L

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<h3>Answer:</h3>

The Equilibrium would shift to produce more NO

<h3>Explanation:</h3>

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N₂(g) + O₂(g) ⇆ 2NO(g)

When a reaction is at equilibrium then the forward reaction rate will be equivalent to the reverse reaction rate. Additionally, the concentration of the reactants and products are the same.
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