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There are 0.94 g mass of [tex]CaCO_{3}[/tex] is required to react completely with 25. 0 ml of 0. 750 m HCl .
Calculation ,
Mass of [tex]CaCO_{3}[/tex] = ?
The 1000ml of HCl = 27.375 g
then the 1 ml of solution contains HCl = 27.375 g/1000×1
25 ml of solution contains HCl = 27.375 g/1000 × 25 = 0.684 g
The chemical equation can be given as :
[tex]CaCO_{3} +HCl[/tex] → [tex]CaCl_{2} +CO_{2} +H_{2} O[/tex]
2 mol of HCl reacts with 1 mol of [tex]CaCO_{3}[/tex]
The amount of [tex]CaCO_{3}[/tex] reacted is given by ,
100/71 × 0.684 g = 0.9639 g = 0.94 g
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