Home
Class 8
CHEMISTRY
When 58 g of Mg(OH)2 reacts with 98 g of...

When 58 g of `Mg(OH)_2` reacts with 98 g of `H_2SO_4`, it gives 36 g of `H_2O` and __________ of `MgSO_4`.

Text Solution

Verified by Experts

The correct Answer is:
120 g
Promotional Banner

Topper's Solved these Questions

  • Language of Chemistry and Transformation of Suubstances

    PEARSON IIT JEE FOUNDATION|Exercise Essay type questions|5 Videos
  • Language of Chemistry and Transformation of Substances

    PEARSON IIT JEE FOUNDATION|Exercise Concept Application Concept Application Level - 3|10 Videos
  • METALS AND NON-METALS

    PEARSON IIT JEE FOUNDATION|Exercise COMPETITION CORNER |64 Videos

Similar Questions

Explore conceptually related problems

Statement 109 % H_(2)SO_(4) represent a way to express concentration of industrial H_(2)SO_(4) . Explanation It represents that 9 g H_(2)O reacts with 40 g SO_(3) to produce 49 g H_(2)SO_(4) in addition to 100 g H_(2)SO_(4) .

A solution of 0.4 g sample of H_(2)O_(2) reacted with 0.632 g of KMnO_(4) in presence of sulphuric acid. The percentage purity of the sample of H_(2)O_(2) is :

Oleum is considered as a solution of SO_(3) in H_(2)SO_(4) , which is obtained by passing SO_(3) in solution of H_(2)SO_(4) When 100 g sample of oleum is diluted with desired mass of H_(2)O then the total mass of H_(2)SO_(4) obtained after dilution is known is known as % labelling in oleum. For example, a oleum bottle labelled as ' 019% H_(2)SO_(4) ' means the 109 g total mass of pure H_(2)SO_(4) will be formed when 100 g of oleum is diluted by 9 g of H_(2)O which combines with all the free SO_(3) present in oleum to form H_(2)SO_(4) as SO_(3)+H_(2)O to H_(2)SO_(4) What is the % of free SO_(3) in an oleum that is labelled as '104.5% H_(2)SO_(4)' ?