5g of an impure sample of CaCo3 reacted with 400cm^3 of 0.2N HCl . The percentage purity of the sample is? Share with your friends Share 2 Yasodharan answered this Dear Student, The computation is expressed below, CaCO3 + 2HCl →CaCl2 + CO2+H2Omolar mass of CaCl2=110.983 gmolmolar mass of HCl = 36.461we have 0.2 N HCl which gives,N=wt in gmeq.wt×V0.2=wt.in gm36.461×0.4⇒wt. in gm = 2.92 g 110.983 g of CaCl2 from 36.461 g of HCltherefore, ⇒2×36.461110.983×2.92=1.92g of pure sample% purity = 1.925×100%=38.4% Regards. 0 View Full Answer G.v Sankirtana answered this experts pls answer my question 2
