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no. of moles of HCl used =mass ×volume ÷1000
=0.75×25÷1000 = 0.01875 = 0.019
no. of mole of CaCO3 to react with HCl =1÷2×no. of moles of HCl ÷1000
=1÷2×0.019÷1000
=9.5 ^-3 .
MASS OF CaCO3 = no. of moles × molar mass
=9.5^-3×100
=9.5 grams
=0.75×25÷1000 = 0.01875 = 0.019
no. of mole of CaCO3 to react with HCl =1÷2×no. of moles of HCl ÷1000
=1÷2×0.019÷1000
=9.5 ^-3 .
MASS OF CaCO3 = no. of moles × molar mass
=9.5^-3×100
=9.5 grams
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1 M HCl = 36.5 g in 1 litre
0.75 M HCl = 27.375 g in 1 litre
For 25 mL = 27.375 x 25 /1000 = 0.68 g HCl
73 g of HCl required to react with 100 gof CaCO3
0.68 g HCl will react with 100 x 0.68 / 73 = 0.93 g CaCO3
Mass of CaCO3 required is 0.93 g
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