How many gram of 90% Na2SO4 can be produced from 250gram of 95% pure NaCl
Answers
(1). When the mixture of sodium carbonate and sodium bicarbonate (2 g) is heated, then sodium bicarbonate decomposes to give sodium carbonate, water and carbon-di-oxide, which are evolved as gases. Therefore the loss in the weight (0.11 g) of the mixture will be due to evolution of theses gases as shown below: 2NaHCO3-Na2Co3+H2O+Co2 62 g (18g+44g) of gas is produced from 168 g of Sodium Bicarbonate. Therefore, 0.11 g of gas will be produced form (0.11x168)/62 = 0.298 g of Sodium Bicarbonate. The weight of sodium bicarbonate is 0.298 g, so, weight of sodium carbonate will be (2g-0.298g) = 1.702g. 2NACL+H2SO4----2HCL+NA2SO4 (2). The given weight of Sodium chloride = 250 g Sodium chloride is 95% pure. Therefore actual weight of Sodium chloride = (95x250)/100 = 237.5 g 117 g of Sodium chloride gives 142 g of Sodium sulphate. Thus, 237.5 g of Sodium chloride will give (237.5x142)/117 = 288.24 g of Sodium sulphate. Now, the sodium sulphate obtained will be 90% pure. Therefore, the actual weight of Sodium sulphate will be = (90x288.24)/100 = 259.46 g of sodium sulphate Thanks & Regards Mukesh Sharma askIITians Faculty