Question

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)​

Answer 1

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