CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

(a) If 112 cm3 of hydrogen sulphide is mixed with 120 cm3 of chlorine at STP, what is the mass of sulphur formed?
H2S + Cl2 2HCl + S
(b) Washing soda has the formula Na2CO3·10H2O. What is the mass of anhydrous sodium carbonate left when all the water of crystallization is expelled by heating 57.2 g of washing soda?
(c) When excess lead nitrate solution was added to a solution of sodium sulphate, 15.1 g of lead sulphate was precipitated. What mass of sodium sulphate was present in the original solution?
Na2SO4 + Pb(NO3)2 PbSO4 + 2NaNO3
(H = 1; C = 12; O = 16; Na = 23; S = 32; Pb = 207)

Open in App
Solution

(a) Moles of hydrogen sulphide in 112 cm3 at STP = Volume of H2SMolar Volume=11222400mol=0.0050
Moles of chlorine in 120 cm3 at STP = Volume of Cl2Molar Volume=12022400mol=0.0053

Hence, hydrogen sulphide is the limiting reagent. Mass of sulphur formed will depend upon the amount of hydrogen sulphide available for the reaction.
According to the balanced chemical equation, one mole of hydrogen sulphide will produce one mole of sulphur.
Thus, 0.005 moles of hydrogen sulphide will produce 0.005 moles of sulphur.
Mass of sulphur formed = Moles of sulphur × Atomic mass of sulphur
= 0.005 × 32
= 0.16 g

(b) Molecular mass of Na2CO3·10H2O = 2(23) + 12 + 3(16) + 20(1) + 10(18) = 286 g
Number of moles in 57.2 g of Na2CO3·10H2O = 57.2286=0.2
Mass of water of crystallisation in 0.2 mol of washing soda = 0.2 × 10(Molecular mass of water)
= 0.2 × 10(18)
= 36 g
Mass of anhydrous sodium carbonate left when all the water of crystallisation is expelled = 57.2 − 36 = 21.2 g

(c) First, let us determine the limiting reagent in the given reaction. As lead nitrate is present in excess; therefore, sodium sulphate is the limiting reagent.
According to the balanced chemical equation, number of moles of lead sulphate produced is equal to number of moles of sodium sulphate used.
Number of moles of lead sulphate formed = Mass of lead sulphateMolecular mass of lead sulphate = 15.1303=0.05 mol
Number of moles of sodium sulphate present in original solution = 0.05 mol
Mass of sodium sulphate present in original solution = Number of moles × Molecular mass of sodium sulphate
= (0.05 × 142) g
= 7.1 g


flag
Suggest Corrections
thumbs-up
0
BNAT
mid-banner-image