2NaCl + H2SO4 ——> Na2SO4 + 2HCl
If 250g NaCl (94.5% pure) is treated with H2SO4, then how much amount of Na2SO4 (83.4% purity) will be obtained?
REFER THIS QUESTION AND TRY THIS BY YOURSELF 2NaCl+H2SO4-----> Na2SO4+2HCl....what mass of 80% pure Na2SO4 can be obtained from 200 kg 90% pure salt?
let x be the mass of Na2SO4
80% of x => 0.8x
moles = 0.8x/142
200 * 1000gm
pure salt = (90/100)*200000 = 180000gm
moles = 180000/58
from eq. concept.
moles * valence factor = constant.
180000/58 * 1 = 0.8x/142 * 2