Question

# $$10$$ L of hard water with temporary hardness of  $$[Ca(HCO_3)_2]$$ requires $$0.56$$ g of lime. The reaction follows as:$$Ca(HCO_3)_2+CaO\longrightarrow 2CaCO_3+H_2O$$.Temporary hardness in terms of ppm of $$CaCO_3$$ is:

A
56 ppm
B
12 ppm
C
100 ppm
D
200 ppm

Solution

## The correct option is D $$200$$ ppm$$1$$ mole of calcium oxide corresponds to $$2$$ moles of calcium carbonate.The molar mass of calcium oxide and calcium carbonate are $$56$$ g/mol and $$100$$ g/mol respectively.$$0.56$$ g $$(0.01$$ mole$$)$$ of lime will correspond to $$2$$ g $$(0.02$$ mole$$)$$ of calcium carbonate.Now, $$10$$ L of water corresponds to $$10000$$ ml.Hence, $$2$$ g calcium carbonate in $$10000$$ ml of water corresponds to $$200$$ ppm.Chemistry

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