Question

If 3.37 grams of calcium nitrate, Ca(NO3)2, react with excess sodium phosphate, Na3PO4, according to 3Ca(NO3)2 + 2Na3PO4 +Ca3(PO4)2 +6NaNO3 what is the theoretical yield of sodium nitrate, NaNO3, in grams? (Enter your answ the hundredth place.)

Answer 1

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 = 3.37 g

mol of Ca(NO3)2 = (mass)/(molar mass)

= 3.37/1.641*10^2

= 2.054*10^-2 mol

According to balanced equation

mol of NaNO3 formed = (6/3)* moles of Ca(NO3)2

= (6/3)*2.054*10^-2

= 4.107*10^-2 mol

Molar mass of NaNO3,

MM = 1*MM(Na) + 1*MM(N) + 3*MM(O)

= 1*22.99 + 1*14.01 + 3*16.0

= 85 g/mol

mass of NaNO3 = number of mol * molar mass

= 4.107*10^-2*85

= 3.491 g

Answer: 3.49 g