in the following chemical reaction between H_2 and Cl_2 to produce HCl, what is the sum of the mass of HCl produced plus the mass of left over reactants when 0.40 g of H_2 completely reacts with 12.35 g of Cl_2?
H_2(g) + Cl_2(g) → 2HCl(g)

Answers

Answer 1

Answer:

Left over mass of hydrogen = 0.06 g

Mass of HCl produced = 12.41 g

Explanation:

Given data:

Mass of H₂ = 0.40 g

Mass of Cl₂ = 12.35 g

Mass of left over reactant = ?

Mass of HCl produced = ?

Solution:

Chemical equation:

H₂ + Cl₂    →     2HCl

Number of moles of H₂:

Number of moles = mass / molar mass

Number of moles = 0.40 g/ 2 g/mol

Number of moles = 0.2 mol

Number of moles of Cl₂:

Number of moles = mass / molar mass

Number of moles = 12.35 g/ 71 g/mol

Number of moles = 0.17 mol

Now we will compare the moles of HCl with H₂ and Cl₂.

                     H₂             :            HCl

                      1               :               2

                   0.2              :           2×0.2 = 0.4

                     Cl₂             :             HCl

                      1               :              2

                    0.17           :            2 × 0.17 = 0.34

Chlorine is limiting reactant.

Mass of HCl produced:

Mass = number of moles × molar mass

Mass = 0.34 mol × 36.5 g/mol

Mass = 12.41 g

Leftover mass of hydrogen:

                          Cl₂               :               H₂

                            1                :                 1

                           0.17            :             0.17

Number of moles of H₂ react with Cl₂  are 0.17.

Moles remain unreacted = 0.2 - 0.17 = 0.03 mol

Mass left over:

Mass = number of moles × molar mass

Mass = 0.03 mol × 2 g/mol

Mass = 0.06 g

Answer 2

The sum of the mass of HCl produced plus the mass of left over reactants is:

Mass of hydrogen = 0.06 g

Mass of HCl  = 12.41 g

Chemical Reaction

Given:

Mass of H₂ = 0.40 g

Mass of Cl₂ = 12.35 g

Mass of left over reactant = ?

Mass of HCl produced = ?

Chemical equation: H₂ + Cl₂    →     2HCl

Number of moles of H₂:

Number of moles = mass / molar mass

Number of moles = 0.40 g/ 2 g/mol

Number of moles = 0.2 mol

Number of moles of Cl₂:

Number of moles = mass / molar mass

Number of moles = 12.35 g/ 71 g/mol

Number of moles = 0.17 mol

The moles of HCl with H₂ and Cl₂.                

  H₂             :            HCl              

     1              :               2          

     0.2          :           2×0.2 = 0.4                

    Cl₂             :             HCl            

        1             :              2          

       0.17         :          2 × 0.17 = 0.34

The chlorine is limiting reactant.

Mass of HCl produced:

Mass = number of moles × molar mass

Mass = 0.34 mol × 36.5 g/mol

Mass = 12.41 g

Leftover mass of hydrogen:                

         Cl₂               :               H₂            

              1             :                 1          

               0.17       :             0.17

Number of moles of H₂ react with Cl₂  are 0.17.

Moles remain unreacted = 0.2 - 0.17 = 0.03 mol

Mass left over:

Mass = number of moles × molar massMass = 0.03 mol × 2 g/molMass = 0.06 g

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Related Questions

A 12 gram piece of Cu at 475 oC is placed in contact with a 15 gram piece of Cr at 265 oC.

Assume the two pieces of metal are the system and that there is no heat exchange between system and surrounding.

What is the final temperature of the two pieces of metal expressed in degree C?

Answers

Answer:

349.22°C

Explanation:

Let the final temperature of the two pieces of metal be x.

Now, the warmer metal which is C u reduces from 475°C to x. Thus Δt for C u is; Δt1 = 475 - x.

The cooler metal Cr increases in temperature from 265°C to x. Thus, it's change in temperature is Δt for Cr is; Δt2 = x - 265.

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