A compound X contains 63.3% manganese and 36.7% oxygen by mass. When X is heated

Question

A compound X contains 63.3% manganese and 36.7% oxygen by mass. When X is heated, oxygen gas is evolved and a new compound Y is containing 72.0% manganese and 28.0% oxygen formed. What are the empirical formulas for compounds X and Y? Write a balanced equation for the conversion of X to Y. A quantity of 25.0 mL of a solution containing both Fe^2+ and Fe^3+ ions is titrated with 23.0 mL of 0.0200 M KMnO_4 in dilute acidic solution. As a result, all of the Fe^2+ ions are converted to Fe^3+ ions. Next, the solution is treated with zinc metal to convert all of the Fe^3+ ions to Fe^2+ ions. Finally, the solution containing only Fe^2+ ions requires 40.0 mL If the same KMnO_4 solution. Calculate the molar concentration of the Fe^2+ and Fe^3+ ions in the original solution. The net ionic equation is: MnO_4 + 5Fe^2+ + 8H^+ rightarrow Mn^2+ + 5Fe^3+ + 4H_2O

Explanation / Answer

For compound X

Mn = 63.3%

O = 36.7 %

Assume that the amount of compound X is 100 g.

So mass of Mn = 63.3 g and O = 36.7 g

Now calculate the moles of Mn and O

Moles of Mn = 63.3 /54.94 = 1.15

Moles of O = 36.7 /15.99 = 2.30

Molar ratio in the compound :

Mn = 1.15/1.15=1

O = 2.30/1.15= 2

Thus the empirical formula of compound X is MnO2

For compound Y

Mn = 72.0%

O = 28.0 %

Assume that the amount of compound X is 100 g.

So mass of Mn = 72.0 g and O = 28.0 g

Now calculate the moles of Mn and O

Moles of Mn = 72.0 /54.94 = 1.31

Moles of O = 28.0 /15.99= 1.75

Molar ratio in the compound :

Mn = 1.31/1.31=1

O = 1.75/1.31= 1.33

Now convert the whole number Mn = 1*3= 3

1.33*3= 4

Thus the empirical formula of compound Y is Mn3O4

The balance reaction for converting X into Y is as follows:

3 MnO2(s) = Mn3O4(s) + O2(g)

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