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CBSE Class 11-science Answered

Calculate the mass of 50% H2SO4 required to decompose 10 g of CaCO3
Asked by aadyaverma700 | 30 Jun, 2021, 04:54: PM
answered-by-expert Expert Answer
This question is based on stoichiometry.
 
Chemical equation for this process can be written as-
C a C O subscript 3 space plus space H subscript 2 S O subscript 4 rightwards arrow space C a S O subscript 4 space plus space H subscript 2 O space plus space C O subscript 2 F r o m space S t o i c h i o m e t r y comma space Mole space ratio space of space straight H subscript 2 SO subscript 4 space and space CaCO subscript 3 space is space 1 colon 1 Molar space mass space of space CaCO subscript 3 space equals 40 plus 12 plus 48 equals 100 Moles space of space CaCO subscript 3 equals 25 over 100 equals 0.25 space moles Since space mole space ratio space is space 1 colon 1 space so space straight H subscript 2 SO subscript 4 space required space will space be equals 0.25 space moles This space equation space is space for space 100 percent sign space straight H subscript 2 SO subscript 4 comma space hence space moles space of space 50 percent sign space straight H subscript 2 SO subscript 4 space required space will space be equals space 0.25 cross times 100 over 50 equals 0.5 space moles Therefore space mass space required space equals space Moles space cross times Molar space mass space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0.5 cross times 100 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 50 space straight g Mass space required space for space 50 percent sign space straight H subscript 2 SO subscript 4 space is space 50 space straight g.
 
Answered by Ravi | 02 Jul, 2021, 06:19: PM

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