Question -
Answer -
0.75 M of HCl тЙб 0.75 mol of HCl are present in 1 L ofwater
тЙб [(0.75 mol) ├Ч (36.5 g molтАУ1)]HCl is present in 1 L of water
тЙб 27.375 g of HCl is present in 1 L of water
Thus, 1000 mL of solution contains 27.375 g of HCl.
Amount of HCl present in 25 mL of solution
= 0.6844 g
From the given chemical equation,
2 mol of HCl (2 ├Ч 36.5 = 73 g) react with 1mol of CaCO3┬а(100 g).
Amount of CaCO3┬аthat willreact with 0.6844 g =
10073├Ч0.6844= 0.9375 g