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andrey2020 [161]
3 years ago
10

How many moles of pcl5 can be produced from 23.0 g of p4 (and excess cl2)?

Chemistry
1 answer:
Masja [62]3 years ago
3 0
The balanced equation for the above reaction is as follows
P4 + 10Cl2 —> 4PCl5
P4 is the limiting reactant. Stoichiometry of P4 to PCl5 is 1:4
The mass of P4 reacted = 23.0 g
Number of P4 moles = 23.0 g/ 124 g/mol = 0.185 mol
According to the stoichiometry the number of PCl5 moles produced are 4 times the amount of P4 moles reacted
Therefore PCl5 moles = 0.185 x 4 = 0.74 mol
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Explanation:

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A tightly sealed 5.0-l flask contains 781 mm hg of ar at 19 °c. the flask is heated until the pressure is doubled. what is the t
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How many calories of heat were added to 347.9 g of water to raise its temperature from 25oC to 55oC?
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10437calories

Explanation:

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2 years ago
Iron(III) oxide and hydrogen react to form iron and water, like this: Fe_2O_3(s) + 3H_2(g) rightarrow 2Fe(s) + 3H_2O(g) At a cer
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Complete Question

The complete question is shown on the first uploaded image

Answer:

The equilibrium constant is  K_c= 2.8*10^{-4}

Explanation:

      From  the question we are told that

              The chemical reaction equation is

      Fe_{2} O_{3}_{(s)} + 3H_{2}_{(g)}  -----> 2Fe_{(s)} + 3H_{2} O_{(g)}

The voume of the misture is  V_m = 5.4L  

  The molar mass of  Fe_{2} O_{3}_{(s)} is a constant with value of  M_{Fe_{2} O_{3}_{(s)} } = 160g/mol

    The molar mass of  H_{2}_{(g)}    is a constant with value of  H_2 = 2g/mol

   

    The molar mass of  H_{2}O    is a constant with value of  H_2O = 18g/mol

Generally the number of moles  is mathematically given as

                     No \ of \ moles \ = \frac{mass}{molar\  mass}

    For   Fe_{2} O_{3}_{(s)}

          No \ of\ moles = \frac{3.54}{160}

                                = 0.022125 \ mols

     For  H_{2}

               No \ of\ moles = \frac{3.63}{2}

                                = 1.815 \ mols

       For  H_{2}O

                         No \ of\ moles = \frac{2.13}{18}

                                              = 0.12 \ mols

Generally the concentration of a compound  is mathematicallyrepresented  as

       Concentration  = \frac{No \ of \ moles }{Volume }

      For   Fe_{2} O_{3}_{(s)}

                Concentration[Fe_2 O_3] = \frac{0.222125}{5.4}

                                         = 4.10*10^{-3}M                          

       For  H_{2}

                  Concentration[H_2] = \frac{1.815}{5.4}

                                           = 0.336M

      For  H_{2}O

                Concentration [H_2O] = \frac{0.12}{5.4}

                                                  = 0.022M

  The equilibrium constant  is mathematically represented as

                K_c = \frac{[concentration \ of \ product]}{[concentration \ of \ reactant ]}

  Considering H_2O  \ for \ product

            And      H_2  \ for  \ reactant

At  equilibrium the

                    K_c = \frac{0.022}{0.336}

                          K_c= 2.8*10^{-4}

3 0
2 years ago
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