Solve number 53 plz 53. The mass of sodium chloride formed when 5.3 g of sodium carbonate is dissolved in 250 ml of 1 2 ? (A) 5.85 g (B) 7.32 g (C) 11.7 g (D) 58.5 g Share with your friends Share 0 Manasa Hegde answered this Dear Student, The equation for the given reaction is Na2CO3 + 2HCl →2NaCl + H2O +CO21 mole of Na2CO3 + 2 mole of HCl gives 2 mole of NaCl 105.98g of Na2CO3 + 73 g of HCl gives 117g of NaCl The amount of product is determined by the limiting reagent. Let us calculate the mass of HCl used in this reaction,Molarity = no. of molesvolume of solution in litre, therefore, no. of moles = M x V = 0.5 x 0.250 = 0.125molesObserve the relation between the no. of moles of sodium carbonate and no. of moles of HCl. No. of moles of sodium carbonate should be half of no. of moles of HCl.no. of moles of Na2CO3=weighed massmolar mass=5.3105.98=0.55 molesThe no. of moles of sodium carbonate is less than hald of no. of moles of HCl, 0.55<12x0.125. Thus sodium carbonate is the limiting reagent.now,105.98g of sodium carbonate gives 117g of sodium chloride, so 5.3g of sodium carbonate gives, 117 x 5.3105.98 = 5.85g of NaClso the answer is option (a) Regards, 1 View Full Answer Jatin answered this 11.7g 1