Question #7953
CaC12(aq)+Na2CO3(aq) > CaCO3(s)+ 2NaC1(aq)
how many moles of sodium carbonate were initially present? 1.06g of sodium carbonate
how many moles of sodium carbonate were initially present? 1.06g of sodium carbonate
Expert's answer
n=m/Mr
n=1.06/(23*2+12+16*3)=1.06/106=0,01 mol
n=1.06/(23*2+12+16*3)=1.06/106=0,01 mol
