First, balancing the equation:
SO_(2(g)) + 2NaOH_((s)) -> Na_2SO_(3(s)) + H_2O_((l))
Calculating moles of each reagent used:
-> We first need to calculate the molar mass of each reagent since we are given the mass of each reagent, and molar mass allows us to convert from mass to moles.
mm(SO_2): 32.07g (molar mass of sulfur) +2(16) (molar mass of oxygen, 2 oxygen atoms present) = 64.07g
n(SO_2) = (53.3cancel(g))/((32.07+2(16))cancel(g)mol^-1)
=0.8319....mol of SO_2
-> Similarly, moles of NaOH:
n(NaOH) = (37.8cancel((g)))/((22.99+16+1.008)cancel(g)mol^-1)
=0.945...mol of NaOH
We know from the balanced equation, that for every one mole of SO_2 which reacts, 2 moles of NaOH are consumed, as the mole ratio of SO_2 to NaOH is 1:2.
Hence, if all 0.8319... moles of SO_2 are consumed, then 2(0.8319...) = 1.6638... moles of NaOH; however, only 0.945...mol of NaOH is present from our calculation above, and thus not enough NaOH is present to react with all the SO_2 present, i.e. NaOH is a limiting reagent.
Since the reaction goes to completion, all NaOH present will thus be consumed (since NaOH is limiting so when all NaOH is consumed, the reaction cannot occur further), and by using the mole ratio from the balanced equation, we can find the moles of each product.
-> Na_2SO_3 : NaOH = 1:2
Therefore moles of Na_2SO_3 = 1/2 moles of NaOH = (0.945...)/2
=0.4725....mol of Na_2SO_3
Mass of Na_2SO_3: using molar mass of Na_2SO_3
0.4725...mol x (2(22.99)+32.07+3(16))gmol^-1
=59.56...g = 59.6g (3 significant figures)
-> H_2O : NaOH = 1:2
Therefore similarly, moles of H_2O = 1/2 moles of NaOH = (0.945...)/2
=0.4725....mol of H_2O
Mass of H_2O:
0.4725...mol x (2(1.008) + 16)gmol^-1
=8.513...g=8.51g (3 significant figures)
