Answer to Question #311610 in General Chemistry for Linda

Question

10 tonnes of limestone is added to 15 tonnes of brine. By using the overall reaction of the process.determine which of the raw materials are in excess.calculate the maximum mass of sodium carbonate that can be produced with these raw materials. Hence calculate the actual mass of sodium carbonate produced if the percentage yield of the process is 84.5%

Answer 1

The preparation of sodium carbonate and sodium bicarbonate from the given raw materials involve solvay process and the reaction involves this process can be given as follows

Solvay Process: - In this reaction carbon dioxide is passed through brine solution ( having 28 % NaCl) and saturated with ammonia to get Sodium carbonate

$NH_3+ H_2O+CO_2 \longrightarrow(NH_4)_2CO_3$

$(NH_4)_2CO_3 +H_2O+CO_2 \longrightarrow2NH_4HCO_3$

$NH_4HCO_3 +NaCl \longrightarrow NaHCO_3 + NH_4Cl$

and then precipitate is filtered and ignited to change into sodium carbonate

$2NaHCO_3 \longrightarrow Na_2CO_3 + CO_2+ H_2O$

and carbon dioxide obtained from limestone as follows

$CaCO_3 \longrightarrow CaO+ CO_2$

Molar mass of limestone = $CaCO_3$ = $40 +12+48 = 100 \frac{g}{mol}$

Mass of limestone = 10 tonnes = 10 $\times$ 1000= 10000 kg

mole of limestone =$\frac{10000}{100}=$ 100000 mole

Mass of brine solution = 15 tonnes = 15000 kg and it has 28 % NaCl

So, mass of NaCl = 4200 Kg

Mole of NaCl = 71794.87 mol

From the above reaction we can find that Indirectly one mole carbon dioxide to reaction with one mole NaCl to produce one mole sodium bicarbonate and 0.5 mole of sodium carbonate

So , here NaCl is in limiting reagent and limestone is in excess state

Maximum mole of Sodium carbonate can be produce = $\frac{71794.87}{2}= 35897.435$

Mass of sodium carbonate = $35897.435 \times 106= 3805128.11 g = 3805.12 kg= 3.805 tonnes$

Mass of actual sodium carbonate = $\frac{3.805 \times 84.5}{100}$ =3.21 tonnes