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A piston containing 0.120moles of methane gas, CH4, has a volume of 2.12liters. If methane is added until the volume is increased to 3.12liters with no change in pressure or temperature, how many grams are in the piston?

Respuesta :

Answer:

2.83 g

Explanation:

At constant temperature and pressure, Using Avogadro's law

[tex]\frac {V_1}{n_1}=\frac {V_2}{n_2}[/tex]

Given ,  

V₁ = 2.12 L

V₂ = 3.12 L

n₁ = 0.120 moles

n₂ = ?

Using above equation as:

[tex]\frac{2.12}{0.120}=\frac{3.12}{n_2}[/tex]

[tex]2.12n_2=0.12\cdot \:3.12[/tex]

[tex]2.12n_2=0.3744[/tex]

[tex]n_2=\frac{0.3744}{2.12}[/tex]

[tex]n_2=0.17660[/tex]

n₂ = 0.17660 moles

Molar mass of methane gas = 16.05 g/mol

So, Mass = Moles*Molar mass = 0.17660 * 16.05 g = 2.83 g

2.83 g  are in the piston.

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