Question

(a) Balance the following equation by entering the smallest possible whole-number coefficients into the blank.

_____ Na₂SO₄(s) + _____C(s) ---- _____ Na₂S(s) + _____ CO(g)

(b) If you start with 49.6 g of sodium sulfate and 16.5 g carbon, calculate the maximum mass of sodium sulfide that could be produced.

_____ g

(c) Assume the reaction trial in part (b) has a 100% yield (or completion rate). Calculate the mass of excess reactant that will remain.

_____ g

(d) If 23.7 g of sodium sulfide is actually produced by the reaction in part (b), calculate the percent yield for this reaction.

_____ %

Answer 1

a) balanced equation is

Na2SO4 + 4C -----------> Na2S + 4CO

49.6g/142g/mol 16.5g/12g/mol 0 0

= 0.3492 =1.375 initial moles

To know the limiting reagent let us find the ratio of moles to required moles

0.3492/1 1.375/4=0.343

Thus carbon is the limiting reagent

  Na2SO4 + 4C -----------> Na2S + 4CO

0.3492 1.375 0 0 initial moles

0.3492 -1.375/4 0 1.375/4 -

= 0.00545 0 0.34375 - after reaction

mass of Na2S produced = moles x molar mass

= 0.34375 molx 78g/mol

= 26.815 g

c)mass of excess reactant = moles x molar mass

= 0.00545 x 142

= 0.7739 g

d) % yield = actual mass x100/ theoretical mass

= 23.7 x100/ 26.815

= 88.38%