Question

What volume of 0.05 mol/L HCl is required to react with 5.00g of manganese dioxide according to this equation?

Answer 1

217 ml of 0.05 M HCl will be required to react with 5 gm of MnO2 according to the equation given.

Explanation:

The balanced chemical equation is given as:

4HCl(aq) + MnO2(s) → 2H2O(l) + MnCl2(aq) + Cl2(g)

This shows that 4 moles of HCl reacts with 1 mole of MnO2

the mass of Manganese oxide is given as 5 grams

molar mass of MnO2 = 86.93 grams/mole

number of moles of MnO2 is given by

number of moles = $\frac{mass}{atomic mass of one mole}$

number of moles= $\frac{5}{86.93}$

                            = 0.0575 moles of MnO2

From the equation:

4 moles of HCl reacts with 1 mole of MnO2

x moles of HCl reacts with $\frac{0.0575}{x}$ moles of MnO2

$\frac{1}{4}$ = $\frac{0.0575}{x}$

= 0.23 moles of HCl will react

atomic mass of HCl = 36.46 Grams/mole

mass = 0.23 x 36.46

         = 8.3858 grams.of HCl

molarity of HCl = $\frac{number of moles}{volume in liters}$

volume is 1 litre

so molarity is 0.23 M

Using the formula

M1V1 = M2V2

0.05 x 1 = 0.23 X x

x = 0.217 litre

so 217 ml of 0.05 M HCl will be required to react with 5 gm of MnO2.