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marusya05 [52]
2 years ago
11

If 0.896 g of a gas occupies a 250 mL flask at 20°C and 760 mm Hg of pressure, what is the molar mass of the gas?​

Chemistry
1 answer:
salantis [7]2 years ago
8 0

Answer:

86.2 g/mol

Explanation:

Before you can find the molar mass, you first need to calculate the number of moles of the gas. To find this value, you need to use the Ideal Gas Law:

PV = nRT

In this equation,

-----> P = pressure (mmHg)

-----> V = volume (L)

-----> n = moles

-----> R = Ideal Gas constant (62.36 L*mmHg/mol*K)

-----> T = temperature (K)

After you convert the volume from mL to L and the temperature from Celsius to Kelvin, you can use the equation to find the moles.

P = 760 mmHg                                      R = 62.36 L*mmHg/mol*K

V = 250 mL / 1,000 = 0.250 L              T = 20 °C + 273.15 = 293.15 K

n = ? moles

PV = nRT

(760 mmHg)(0.250 L) = n(62.36 L*mmHg/mol*K)(293.15 K)

190 = n(18280.834)

0.0104 = n

The molar mass represents the mass (g) of the gas per every 1 mole. Since you have been given a mass and mole value, you can set up a proportion to determine the molar mass.

\frac{?grams}{1 mole} =\frac{0.896grams}{0.0104moles}                                      <----- Proportion

?grams(0.0104moles) = 0.896                       <----- Cross-multiply

?grams = 86.2                                               <----- Divide both sides by 0.0104

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Answer :  The amount of reactant left in excess is 13.1075 grams.

Explanation : Given,

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Molar mass of NaOH = 40 g/mole

First we have to calculate the moles of NH_4Cl and NaOH.

\text{Moles of }NH_4Cl=\frac{\text{Mass of }NH_4Cl}{\text{Molar mass of }NH_4Cl}=\frac{26.5g}{53.5g/mole}=0.495moles

\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{10g}{40g/mole}=0.25moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

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From the balanced reaction we conclude that

As, 1 mole of NaOH react with 1 mole of NH_4Cl

So, 0.25 moles of NaOH react with 0.25 moles of NH_4Cl

From this we conclude that, NH_4Cl is an excess reagent because the given moles are greater than the required moles and NaOH is a limiting reagent and it limits the formation of product.

Moles of remaining excess reactant = 0.495 - 0.25 = 0.245 moles

Now we have to calculate the mass of NH_4Cl.

\text{Mass of }NH_4Cl=\text{Moles of }NH_4Cl\times \text{Molar mass of }NH_4Cl

\text{Mass of }NH_4Cl=(0.245mole)\times (53.5g/mole)=13.1075g

Therefore, the amount of reactant left in excess is 13.1075 grams.

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