How much mass of caco3 is required to react completely with 100ml of 0.5m hcl?
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Caco3 + 2 hcl---> cacl2 + co2 + h2o
0.5M= no of moles/ 1 litre
Therefore for 1 L there are 0.5 moles of hcl
For 100 ml, we get 0.5 x 100/1000= 0.05 mol hcl
Now from equation 1 Mol caco3 reacts with 2 mol hcl
So x mol caco3 reacts with 0.05 mol hcl
By cross multiplication we get 0.05/2 = 1/40 mol caco3
1/40mol= some mass of caco3/ 100(molar mass of caco3)
Mass of caco3= 1/40 x 100= 2.5 g
0.5M= no of moles/ 1 litre
Therefore for 1 L there are 0.5 moles of hcl
For 100 ml, we get 0.5 x 100/1000= 0.05 mol hcl
Now from equation 1 Mol caco3 reacts with 2 mol hcl
So x mol caco3 reacts with 0.05 mol hcl
By cross multiplication we get 0.05/2 = 1/40 mol caco3
1/40mol= some mass of caco3/ 100(molar mass of caco3)
Mass of caco3= 1/40 x 100= 2.5 g
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