Question

An 11.2 L sample of gas is determined to contain 0.5 moles of nitrogen. At the same temperature and pressure, how many moles of gas would there be in a 20 L sample

Answer 1

Answer:

0.89 moles of N2 gas

Explanation:

From Avogadro's law:

V1/n1 = V2/n2

n2 = V2n1/V1

V1=11.2L

V2= 20L

n1=0.5 moles

n2 = ??

Therefore

n2= 20×0.5/11.2= 0.89 moles

Answer 2

Answer:

0.8929 mol

Explanation:

An 11.2-L sample of gas is determined to contain .50 mol N2. at the same temperature and pressure, how many moles of gas would there be in a 20. -L sample?

According to avogadro's law

$\frac{V_1}{n_1} = \frac{V_2}{n_2}$

From the given data

V1 = 11.2L

n1 = 0.5mole

V2 = 20L

n2 =?

$\frac{11.2}{0.5} = \frac{20}{?}$

$22.4=\frac{20}{?} \\\\= \frac{20}{24} \\\\= 0.8929$

= 0.8929 mol