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gregori [183]
3 years ago
15

A 1.25 g gas sample occupies 663 ml at 25∘ c and 1.00 atm. what is the molar mass of the gas?

Chemistry
1 answer:
lakkis [162]3 years ago
3 0

Using ideal gas equation,  

P\times V=n\times R\times T

Here,  

P denotes pressure  

V denotes volume  

n denotes number of moles of gas  

R denotes gas constant  

T denotes temperature  

The values at STP will be:  

P=1 atm  

T=25 C+273 K =298.15K

V=663 ml=0.663L

R=0.0821 atm L mol ⁻¹

Mass of gas given=1.25 g g

Molar mass of gas given=?

Number of moles of gas, n= \frac{Given mass of the gas}{Molar mass of the gas}

Number of moles of gas, n= \frac{1.25}{Molar mass of the gas}

Putting all the values in the above equation,

1\times 0.663=\frac{1.25}{Molar mass of the gas}\times 0.0821\times 298.15

Molar mass of the gas=46.15

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Based on the equation, how many grams of Br2 are required to react completely with 36.2 grams of AlCl3?
s2008m [1.1K]

Answer:

65.08 g.

Explanation:

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<em>2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,</em>

2.0 mole of AlCl₃ reacts with 3.0 mole of Br₂ to produce 2.0 mole of AlBr₃ and 3.0 mole of Cl₂.

  • Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:

<em>n = mass/molar mass</em> = (36.2 g)/(133.34 g/mol) = <em>0.2715 mol.</em>

<u><em>Using cross multiplication:</em></u>

2.0 mole of AlCl₃ reacts with → 3.0 mole of Br₂, from the stichiometry.

0.2715 mol of AlCl₃ reacts with → ??? mole of Br₂.

∴ The no. of moles of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃  = (0.2715 mol)(3.0 mole)/(2.0 mole) = 0.4072 mol.

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