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5m hcl in 50ml reacts with 5g mno₂. calculate the mass of mncl₂ and volume of cl₂.

The answer to your question is:

a) 1.12 l of Cl2

b) 6.3 g of MnCl2

Explanation:

Reaction

                   4HCl + MnO2 ⇒  Cl2 + 2H2O + MnCl2

Data

HCl 5M, 50 ml (0.05 l)

MnO2 = 5 g

MW MnO2 = 87 g

MW MnCl2 = 126 g

Process

For HCl

M = n/V

n = MV

n = 5(0.05) = 0.25 moles     from the reaction 4(0.05) = 1 mol of HCl

For MnO2

                        87 g 1 mol

                           5 g   x

                         x = 5/87 = 0.05 mol of MnO2

The proportion in the reaction HCl to MnO2 is 4:1 but from the operations,  

1: 0.05, the from this we see that MnO2 is the limiting reactant, and we take MnO2 to obtain the values of the questions.

a)  From the reaction        1 mol of MnO2 1 mol of Cl2

                                          0.05 mol               x

                                          x = 0.05 moles of Cl2

                        1 mol of Cl2  22.4 l

                        0.05 moles       x

           x = 0.05 x 22.4/1 = 1.12 l

b) From the reaction        1 mol of MnO2 1 mol of MnCl2

                                        0.05 mol of MnO2   x

                                        x = 0.05 moles of MnCl2

                                        1 mol of MnCl2  126 g of MnCl2

                                       0.05 mol of MnCl2    x

                        x = 0.05 x 126 = 6.3 g

d: the voltage of the secondary coil is less than that of the primary coil.

im not suree sorry if wrong