Answer: 67.8 %.
Explanation:
Okay, let us delve right into the solution to the question;
The balanced chemical reaction is given by the equation (1) below;
2 HCOONa + H2SO4 ---------> 2 CO + 2 H2O + Na2SO4. ----------------------------------------------------------------------------(1).
From the balanced chemical reaction in equation (1) above we can see that; 2 moles of HCOONa reacts with one moles of tetraoxosulphate acid, H2SO4 to produce 2 moles of carbonmonoxide,CO; 2 moles of water, H2O and 1 mole of sodium tetraoxosulphate, Na2SO4.
The parameters given from the question are; total atmospheric pressure, P(t) = 752 torr, volume of CO= 242 mL = 0.242 Litres.
STEP ONE : find the carbon monoxide,CO pressure; P(CO).
Using the formula below;
P(t) = P(CO) + P(H2O). Hence;
P(CO) = P(t) - P(H2O). Note that P(H2O)= 19.8 torr.
==>P(CO)= 752 torr - 19.8 torr = 732.2 torr.
STEP TWO: calculate the number of moles of Carbonmonoxide,CO.
Using the formula below;
Number of moles= pressure(P) × volume(v) / gas constant(R) × temperature (T).
That is, n= PV/RT.
n= 732 torr × 0.242 Litres/ 62.4 × 295.15.
= 9.62 × 10^-3 mol of CO.
STEP THREE:
2 moles of HCOONa = 2 moles of CO.
=> 2 moles of HCOONa = 2 moles of CO/ 2 moles of CO = 1 mol( HCOONa/ CO).
Then, 9.62 × 10^-3 mol of CO × 1 mol( HCOONa/ CO).
==> 9.62 × 10^-3 mol HCOONa × molar masss of HCOONa(68 grams/mol)
= 0.654 grams.
Therefore, the percentage of sodium formate in the original mixture = 0.654 grams/ 0.964 gram × 100 = 67.8 %.
