Question

A solution of 0.4g sample of H2O2 reacted with 0.632g of KMnO4 in presence of sulphuric acid. Calculate percentage purity of sample of H2O2.

Answer 1

Answer:

The Percentage Purity of H2O2 is 17%.

Explanation:  

Chemcial Reaction can be written as:

$KMnO_{4}+H_{2}O_{2}=2KOH+MnO_{2}+O_{2}\\According \ to \ the \ Reaction \\\2 \ Moles \ of \ KMnO_{4} \ reacts \ with = 1 Mole \ of \ H_{2}O_{2}\\158*2g \ of \ KMnO_{4}\ reacts \ with \ H_{2}O_{2}=34*1g\\ 1g \ of \ KMnO_{4}\ reacts \ with \ H_{2}O_{2}=(\frac{34}{152*2} )g\\0.632g \ of \ KMnO_{4}\ reacts \ with \ H_{2}O_{2}=0.632*(\frac{34}{152*2} )g\\0.632g \ of \ KMnO_{4}\ reacts \ with \ H_{2}O_{2}=0.068g$

Given, H2O2 used in reaction is=0.4g

so, Percentage Purity can be calculated as:

%age Purity=(Amount used in reaction /Total amount of sample )*100

$=(\frac{0.068}{0.4})*100\\= 17 \%$

Hope You find It helpful  :)

Answer 2

Answer:

above are the answer of question