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bixtya [17]
1 year ago
9

An unknown gas effuses at a rate of 2.0 times the rate of Cl₂. What is the molar mass of the unknown gas?

Chemistry
1 answer:
Ratling [72]1 year ago
3 0

The molar mass of the unknown gas is determined as 17.75 g.

<h3>Rate of gas diffusion</h3>

The rate of gas diffusion is given by Graham's law.

R₁√M₁ = R₂√M₂

where;

  • R₁ is the rate of diffusion of the unknown gas
  • M₁ is the molecular mass of the unknown gas
  • R₂ is rate of diffusion of chlorine gas
  • M₂ is the molecular mass of chlorine gas

\frac{R_1}{R_2} = \sqrt{\frac{M_2}{M_1} } \\\\\frac{2R_2}{R_2} = \sqrt{\frac{71}{M_1} }\\\\2 = \sqrt{\frac{71}{M_1} }\\\\2^2 = \frac{71}{M_1} \\\\M_1 = \frac{71}{4} \\\\M_1 = 17.75 \ g

Learn more about rate of diffusion here:  brainly.com/question/11197959

#SPJ1

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For the following reaction, 33.7 grams of bromine are allowed to react with 13.0 grams of chlorine gas.
Ira Lisetskai [31]

Explanation:

The reaction is given as;

Br2(g) + Cl2(g) ----> 2BrCl(g)

From the equation;

1 mol of Br2 reacts with 1 mol of Cl2

Converting the masses given to moles, using the formular;

Number of moles = Mass / Molar mass

Br2;

Number of moles = 33.7 g / 159.808 g/mol = 0.21088 mol

Cl2;

Number of moles = 13.0 g / 70.906 g/mol = 0.18334 mol

From the values;

0.18334 mol of Cl2 would react with 0.18334 mol of Br2 with an excess of 0.02754 mol of Br2

<em>What is the maximum amount of bromine monochloride that can be formed? __________grams</em>

<em />

1 mol of Cl2 produces 2 mol of Bromine Monochloride

0.18334 mol of Cl2 would produce x

Solving for x;

x = 0.18334 * 2 = 0.36668 mol

Converting to mass;

Mass = Number of moles * Molar mass = 0.36668 mol * 115.357 g/mol

Mass = 42.299 g

<em />

<em>What is the FORMULA for the limiting reagent?</em>

<em />

The limiting reagent is Cl2 as it determines the amount of product formed. The moment the reaction uses up Cl2, the reaction stops.

<em>What amount of the excess reagent remains after the reaction is complete? __________grams</em>

The excess reagent is Br2

The number of moles left is;

0.02754 mol of Br2

Converting to mass;

Mass = Number of moles * Molar mass = 0.02754 mol  * 159.808 g/mol

Mass = 4.401 g

7 0
2 years ago
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