Question

Question 12 hydrogen gas has a density of 0.090/gl , and at normal pressure and 2.57°c one mole of it takes up 22.4l . how would you calculate the moles in 410.g of hydrogen gas?

Answer 1

Answer : The moles of hydrogen gas will be, 201.9 moles

Solution :

First we have to calculate the molar mass of hydrogen gas.

using ideal gas equation,

$PV=nRT\\\\P=\frac{w}{V}\times \frac{RT}{M}\\\\P=D\times \frac{RT}{M}$

where,

n = number of moles of gas

w = mass of gas

P = pressure of the gas = 1 atm

T = temperature of the gas = $2.57^oC=273+2.57=275.57K$

M = molar mass of hydrogen gas = ?

R = gas constant = 0.0821 Latm/moleK

D = density of gas  = 0.090 g/L

Now put all the given values in the above equation, we get the molar mass of hydrogen gas.

$1atm=0.090g/L\times \frac{(0.0821L.atm/mole.K)\times (275.57K)}{M}$

$M=2.03g/mole$

Now we have to calculate the moles of hydrogen gas.

$\text{Moles of }H_2=\frac{\text{Mass of }H_2}{\text{Molar mass of }H_2}=\frac{410g}{2.03g/mole}=201.9mole$

Therefore, the moles of hydrogen gas is, 201.9 moles

Answer 2

The moles in 410 g of hydrogen gas : 201.67

Further explanation

Some of the laws regarding gas can apply to ideal gas (volume expansion does not occur when the gas is heated), :

Boyle's law at constant T, $\displaystyle P=\frac{1}{V}$

Charles's law, at constant P, $\displaystyle V=T$

Avogadro's law, at constant P and T, $\displaystyle V=n$

So that the three laws can be combined into a single gas equation, the ideal gas equation

In general, the gas equation can be written

$\large{\boxed{\bold{PV=nRT}}}$

where

P = pressure, atm

V = volume, liter

n = number of moles

R = gas constant = 0.082 l.atm / mol K

T = temperature, Kelvin

Hydrogen gas has a density of 0.090, and at normal pressure (1 atm) and 2.57 °C (273+ 2.57 = 275.57K), one mole of it takes up 22.4 l

From here we can input into the ideal gas equation to find an unknown value, molar mass (M)

we have to manipulate this ideal gas equation to find molar mass,

molar mass is the ratio between mass and mole

$\displaystyle M=\frac{m}{n}$.... equation 1

Whereas density is the ratio between  mass and volume

$\displaystyle \rho=\frac{m}{v}$.... equation 2

The ideal gas equation becomes

$\displaystyle PV=\frac{M}{m}RT$

$\displaystyle M=\frac{m}{v}\frac{RT}{P}$

$\displaystyle M=\rho\frac{RT}{P}$

We input the known values

$\displaystyle M=0.090\frac{0.082.275.57}{1}$

$\large{\boxed{\bold{M=2.033}}}$

moles in 410 g of hydrogen gas :

$\displaystyle n=\frac{m}{M}$

$\displaystyle n=\frac{410}{2.033}$

$\large{\boxed{\bold{n=201.67}}}$

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Keywords: mole, hydrogen gas, density, normal pressure