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`80%` शुद्धता वाले `KNO_(3)` के 50 ग्राम से नाइट्रिक अम्ल के कितने ग्राम प्राप्त किये जा सकते है ?

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Correct Answer - 24.9 ग्राम
संगत समीकरण है- `underset(2xx101.11"ग्राम")(2KNO_(3))+H_(2)SO_(4)tounderset(2xx63.018"ग्राम")(2HNO_(3))+K_(2)SO_(4)`
चूँकि दिया गया नमूना `80%` शुद्ध है अतः इसके `50.0` ग्राम में शुद्ध `KNO_(3)` की मात्रा `=(50.0xx80)/(100)=40.0` ग्राम
`because 2xx101.11`ग्राम `KNO_(3)`से प्राप्त `HNO_(3)=2xx63.018`
`therefore40.0` ग्राम `KNO_(3)` से प्राप्त ` HNO_(3)=(2xx63.018)/(2xx101.11)xx40.0=24.9` ग्राम
अतः `KNO_(3)`का दिया गया नमूना `24.9`ग्राम `HNO_(3)`दे सकता है.

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