Question

How many grams of KCl 03 are needed to produce 6.75 Liters of O2 gas measured at 1.3 atm pressure and 298 K?

Answer 1

11.48-gram of $KCl0_3$ are needed to produce 6.75 Liters of $O_2$  gas measured at 1.3 atm pressure and 298 K

What is an ideal gas equation?

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

First, calculate the moles of the gas using the gas law,

PV=nRT, where n is the moles and R is the gas constant. Then divide the given mass by the number of moles to get molar mass.

Given data:

P= 1.3 atm

V= 6.75 Liters

n=?

R= $0.082057338 \;L \;atm \;K^{-1}mol^{-1}$

T=298 K

Putting value in the given equation:

$\frac{PV}{RT}=n$

$n= \frac{1.3 \;atm\; X \;6.75 \;L}{0.082057338 \;L \;atm \;K^{-1}mol^{-1} X 298}$

Moles = 0.3588 moles

Now,

$Moles = \frac{mass}{molar \;mass}$

$0.3588 moles = \frac{mass}{32}$

Mass= 11.48 gram

Hence, 11.48-gram of $KCl0_3$ are needed to produce 6.75 Liters of $O_2$ gas measured at 1.3 atm pressure and 298 K

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