Question

what volume of 0.550 mol/L sodium carbonate, would be required to react with 200 mL of a 0.393 mol/L Cu(NO₃)₂ solution?

Answer 1

    Na2CO3 + Cu(NO3)2 -------------- 2NaNO3 + CuCO3

   1 mole              1 mole

Na2CO3 = 0.550 M

Cu(NO3)2 = 200 mL of 0.393M

number of moles of Cu(NO3)2 = 0.393Mx0.200L = 0.0786 moles

according to equation

1 mole of Cu(NO3)2 = 1 mole of Na2CO3

0.0786 moles of Cu(NO3)2 = 0.0786 moles of Na2CO3

number of moles of Na2CO3 = 0.0786 moles

Molarity of Na2CO3 = number of moles/volume

Volume = number of moles/Molarity

Volume = 0.0786/0.550 =0.142L

Volume = 0.142L = 0.142x1000= 142mL

Volume of Na2CO3 = 142 mL.