Answer to Question #58001 in Other Chemistry for Khaled
what is the mass of sodium carbonate (eq.wt id na2co3=53 ) which should react with 25 ml of 1N hcl?
25 ml of 1N HCl contains the following numbers of mole of HCl: ν(HCl) = C(HCl)×V(HCl) = 0.025 L × 1N = 0.025 mol Since sodium carbonate reacts with two molecules of HCl: Na2CO3 + 2HCl → 2NaCl + H2CO3 The amount of sodium carbonate is ν(HCl)/2 or 0.025 mol/2 = 0.0125 mol. The mass of Na2CO3 equals: m = ν(Na2CO3)×Mr(Na2CO3), where Mr – the molecular mass. Hence, the mass is: m = 0.0125 mol × 106 g/mol = 1.325 g.