Question

For the reaction

Na₂CO₃+Ca(NO₃)₂ → CaCO₃+2NaNO₃

how many grams of calcium nitrate, Ca(NO₃)₂, are needed to react completely with 40.5 g of sodium carbonate, Na₂CO₃ ?

Answer 1

Molar mass of Na2CO3,

MM = 2*MM(Na) + 1*MM(C) + 3*MM(O)

= 2*22.99 + 1*12.01 + 3*16.0

= 105.99 g/mol

mass of Na2CO3 = 40.5 g

mol of Na2CO3 = (mass)/(molar mass)

= 40.5/1.06*10^2

= 0.3821 mol

According to balanced equation

mol of Ca(NO3)2 reacted = moles of Na2CO3

= 0.3821 mol

Molar mass of Ca(NO3)2,

MM = 1*MM(Ca) + 2*MM(N) + 6*MM(O)

= 1*40.08 + 2*14.01 + 6*16.0

= 164.1 g/mol

mass of Ca(NO3)2 = number of mol * molar mass

= 0.3821*1.641*10^2

= 62.7 g

Answer: 62.7 g