Question

If a solution containing 42.33 g of mercury(II) chlorate is allowed to react completely with a solution containing 6.256 g of sodium sulfide, how many grams of solid precipitate will form? How many grams of the reactant in excess will remain after the reaction?

Answer 1

1)

Molar mass of Hg(ClO3)2,

MM = 1*MM(Hg) + 2*MM(Cl) + 6*MM(O)

= 1*200.6 + 2*35.45 + 6*16.0

= 367.5 g/mol

mass(Hg(ClO3)2)= 18.15 g

use:

number of mol of Hg(ClO3)2,

n = mass of Hg(ClO3)2/molar mass of Hg(ClO3)2

=(18.15 g)/(3.675*10^2 g/mol)

= 4.939*10^-2 mol

Molar mass of Na2S,

MM = 2*MM(Na) + 1*MM(S)

= 2*22.99 + 1*32.07

= 78.05 g/mol

mass(Na2S)= 20.0 g

use:

number of mol of Na2S,

n = mass of Na2S/molar mass of Na2S

=(20 g)/(78.05 g/mol)

= 0.2562 mol

Balanced chemical equation is:

Hg(ClO3)2 + Na2S ---> HgS + 2 NaClO3

1 mol of Hg(ClO3)2 reacts with 1 mol of Na2S

for 4.939*10^-2 mol of Hg(ClO3)2, 4.939*10^-2 mol of Na2S is required

But we have 0.2562 mol of Na2S

so, Hg(ClO3)2 is limiting reagent

we will use Hg(ClO3)2 in further calculation

Molar mass of HgS,

MM = 1*MM(Hg) + 1*MM(S)

= 1*200.6 + 1*32.07

= 232.67 g/mol

According to balanced equation

mol of HgS formed = (1/1)* moles of Hg(ClO3)2

= (1/1)*4.939*10^-2

= 4.939*10^-2 mol

use:

mass of HgS = number of mol * molar mass

= 4.939*10^-2*2.327*10^2

= 11.49 g

Answer: 11.5 g

2)

According to balanced equation

mol of Na2S reacted = (1/1)* moles of Hg(ClO3)2

= (1/1)*4.939*10^-2

= 4.939*10^-2 mol

mol of Na2S remaining = mol initially present - mol reacted

mol of Na2S remaining = 0.2562 - 4.939*10^-2

mol of Na2S remaining = 0.2069 mol

use:

mass of Na2S,

m = number of mol * molar mass

= 0.2069 mol * 78.05 g/mol

= 16.15 g

Answer: 16.2 g