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Rashid [163]
3 years ago
12

Determine the molecular mass of a gas where 3.87 g occupies 0.896 l at standard conditions.

Chemistry
1 answer:
Arturiano [62]3 years ago
5 0
Let's assume that the gas has ideal gas behavior.

Ideal gas law,
<span>
PV = nRT               (1)

Where, P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant ( 8.314 J mol</span>⁻¹ K⁻<span>¹) and T is temperature in Kelvin.

</span>n = m/M          (2)

Where, n is number of moles, m is mass and M is molar mass.

From (1) and (2),
PV = (m/M) RT

By rearranging,
M = (mRT)/PV              (3)

P =  standard pressure = 1 atm = 101325 pa
V = 0.896 L = 0.896 x 10⁻³ m³
R = 8.314 J mol⁻¹ K⁻¹<span>
T = Standard temperature = 273 K
m = </span>3.87 g = 3.87 x 10⁻³ kg<span> 
M = ?
</span><span>
By appying the formula,
M =(</span>3.87 x 10⁻³ kg x 8.314 J mol⁻¹ K⁻¹ x 273 K) /101325 pa x  0.896 x 10⁻³m³
<span>M = 0.0967 kg
M = 96.7 g.

Hence, the molar mass of the gas is 96.7 g.

</span>
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15.00 grams of Chromium react with 15.00 grams of hydrobromic acid. Calculate the theoretical yield of the reaction. At STP what
s2008m [1.1K]

Answer:

(a) 18.03 g

(b) 2.105 L

(c) 85.15 %

Step-by-step explanation:

We have the masses of two reactants, so this is a<em> limiting reactant problem.  </em>

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.  

<em>Step 1</em>. <em>Gather all the information</em> in one place with molar masses above the formulas and masses below them.  

M_r:        52.00   80.91       291.71

                2Cr  +  6HBr ⟶ 2CrBr₃ + 3H₂

Mass/g:  15.00    15.00  

<em>Step 2</em>. Calculate the <em>moles of each reactant</em>  

  Moles of Cr = 15.00 × 1/52.00

  Moles of Cr = 0.2885 mol Cr

Moles of HBr = 15.00 × 1/80.91

Moles of HBr = 0.1854 mol HBr ×  

<em>Step 3</em>. Identify the<em> limiting reactant</em>  

Calculate the moles of CrCl₃ we can obtain from each reactant.  

<em>From Cr</em>:

The molar ratio of CrBr₃:Cr is 2 mol CrBr₃:2 mol Cr

Moles of CrBr₃ = 0.2885 × 2/2

Moles of CrBr₃ = 0.2885 mol CrCl₃

<em>From HBr: </em>

The molar ratio of CrBr₃:HBr is 2 mol CrBr₃:6 mol HBr.

Moles of CrBr₃ = 0.1854 × 2/6

Moles of CrBr₃ = 0.061 80 mol CrBr₃

The limiting reactant is HBr because it gives the smaller amount of CrBr₃.

<em>Step 4</em>. Calculate the <em>theoretical yields</em> of CrBr₃ and H₂.

Theoretical yield of CrBr₃ = 0.061 80 × 291.71/1

Theoretical yield of CrBr₃ = 18.03 g CrCl₃

The molar ratio is 3 mol H₂:6 mol HBr

   Theoretical yield of H₂ = 0.1854 × 3/6

   Theoretical yield of H₂ = 0.092 70 mol H₂

<em>Step 5</em>. Calculate the <em>volume of H₂</em> at STP

STP is 1 bar and 0 °C.

The molar volume of a gas at STP is 22.71 L.

Volume = 0.092 70 × 22.71/1

Volume = 2.105 L

<em>Step 6</em>. Calculate the <em>percent yield </em>

       % Yield = actual yield/theoretical yield × 100 %

Actual yield = 15.35 g

       % yield = 15.35/18.03 × 100

       % yield = <em>85.15 % </em>

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