If a solution containing 102.78 gof mercury(II) nitrate is allowed to react completely with a solution containing 17.796 g of sodium sulfide, how many grams of solid precipitate will form?

53 g of HgS will be formed

Explanation:

We determine the main reaction:

Hg(NO₃)₂(aq) + Na₂S (aq) → HgS (s)↓ + 2NaNO₃(aq)

We determine the limiting reactant (we need to convert the mass to moles, firstly)

102.78 g / 324.59 g/mol = 0.316 moles of nitrate

17.796 g / 78.06 = 0.228 moles of sulfide

Ratio is 1:1, so the limiting reactant is the Na₂S. For 0.316 moles of nitrate, I need the same amount of sulfide and I only have 0.228 moles.

Ratio is again 1:1. For 0.228 moles of sodium sulfide I can produce the same amount of mercury(II) sulfide

We convert the moles to mass → 232.65 g/mol . 0.228 mol = 53 g of HgS will be formed

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml