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kirill [66]
3 years ago
10

A 15.0-L scuba diving tank contains a helium-oxygen (heliox) mixture made up of 23.0 g of He and 4.14 g of O2 at 298 K. Calculat

e the mole fraction of each component in the mixture.
Chemistry
1 answer:
EleoNora [17]3 years ago
3 0

<u>Answer:</u> The mole fraction of Helium is 0.978 and that of oxygen gas is 0.022

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

  • <u>For helium:</u>

Given mass of helium = 23 g

Molar mass of helium = 4 g/mol

Putting values in equation 1, we get:

\text{Moles of helium}=\frac{23g}{4g/mol}=5.75mol

  • <u>For oxygen gas:</u>

Given mass of oxygen gas = 4.14 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=\frac{4.14g}{32g/mol}=0.129mol

Mole fraction of a substance is given by:

\chi_A=\frac{n_A}{n_A+n_B}

  • <u>For helium:</u>

Moles of helium = 5.75 moles

Total moles = [5.75 + 0.129] = 5.879 moles

Putting values in above equation, we get:

\chi_{(He)}=\frac{5.75}{5.879}=0.978

  • <u>For oxygen gas:</u>

Moles of oxygen gas = 0.129 moles

Total moles = [5.75 + 0.129] = 5.879 moles

Putting values in above equation, we get:

\chi_{(O_2)}=\frac{0.129}{5.879}=0.022

Hence, the mole fraction of Helium is 0.978 and that of oxygen gas is 0.022

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n(NH3)=m(NH3)/M(NH3)=135700/17,03=7968.29 mol

n(CO2)=m(CO2)/M(CO2)=211400/44.01=4803.45 mol

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Answer:

KBr is limiting reactant.

Explanation:

Given data:

Mass of  KBr =4g

Mass of Cl₂ = 6 g

Limiting reactant = ?

Solution:

Chemical equation:

2KBr + Cl₂      →    2KCl + Br₂

Number of moles of KBr:

Number of moles = mass/molar mass

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Number of moles = mass/molar mass

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Number of moles = 0.09 mol

Now we will compare the moles of reactant with product.

              KBr            :            KCl

                2              :              2

            0.03            :            0.03

             KBr            :              Br₂

                2             :               1

             0.03           :          1/2×0.03= 0.015

               Cl₂             :            KCl

                 1              :              2

            0.09            :           2/1×0.09 = 0.18

               Cl₂             :              Br₂

                1              :               1

             0.09           :            0.09

Less number of moles of product are formed by the KBr thus it will act as limiting reactant while Cl₂  is present in excess.

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