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50g of caco3 is allowed to react with 73.5g of h3po4 calculate 1 amount of ca3(po4)2 formed in moles and 2. Amount of unreacted reagents in moled
Dear Nancy in my point of veiw,balanced equation- 3CaCo3+2H2Po4 gives Ca3(Po4)2+3Co2+3H2O so210g of CaCo3 and 196 H3Po4 gives 310g of Ca3(Po4)According to problem,50gof CaCo3 and 73.5 gH3Po4 gives x g of Ca3(pO4) then ‘x=(50+73.5)*310/210+196i.e. equal to 75.66205g RegardsArun
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