Question

1)Consider the combustion of carbon monoxide (CO) in oxygen gas: 2CO(g) + O₂(g→2CO₂(g) Starting with 3.82 moles of CO, calculate the number of moles of CO₂ produced if there is enough oxygen gas to react with all the CO. Answer in mol CO₂

₂₎a) What is the Na⁺ concentration in each of the following solutions: 3.65 M sodium sulfate:2.18 M sodium carbonate:0.385 M sodium bicarbonate: What is the concentration of a lithium carbonate solution that is 0.395 M in Li⁺? All answer in M

3)A sample of 0.7960 g of an unknown compound containing barium ions (Ba²⁺) is dissolved in water and treated with an excess of Na₂SO₄. If the mass of the BaSO₄ precipitate formed is 0.7272 g, what is the percent by mass of Ba in the original unknown compound?

4)A volume of 50.2 mL of a 0.168 M Ca(NO₃)₂ solution is mixed with 80.5 mL of a1.496 M Ca(NO₃)₂ solution. Calculate the concentration of the final solution.Answer in M

Answer 1

1.

2 CO (g) + O2 (g) ------------> 2 CO2 (g)

From the baoanced equation,

2 mol of CO produces 2 mol of CO2

3.82 mol of CO produces 3.82 mol of CO2

SO, moles of CO2 produced = 3.82 mol

(2)

(a)

Na2CO3 (aq.) ---------------> 2 Na+ (aq.) + CO32- (aq.)

[Na+] = 2 [Na2CO3] = 2 * 3.65 = 7.30 M

(b)

Na2SO4 (aq.) -----------> 2 Na+ (aq.) + SO42- (aq.)

[Na+] = 2 * [Na2SO4] = 2 * 2.18 = 4.36 M

(c)

NaHCO3 (aq.) -------------> Na+ (aq.) + HCO3- (aq.)

[Na+] = [NaHCO3] = 0.385 M

(d)

Li2CO3 (aq.) ------------> 2 Li+ (aq.) + CO32- (aq.)

[Li+] = 2 [Li2CO3] = 2 * 0.395 = 0.790 M

(3)

Na2SO4 (aq.) + Ba2+ (aq.) ------------> BaSO4 (s) + 2 Na+ (aq.)

Moles of BaSO4 = mass / molar mass = 0.7272 / 233.4 = 0.00312 mol

Moles of Ba2+ = Moles of BaSO4 = 0.00312 mol

Mass of Ba = 0.00312 * 137.3 = 0.4284 g.

% of Ba in the sample = 0.4284 * 100 / 0.7960 = 53.8 %

(4)

Final concentration = [(0.168*50.2) + (1.496*80.5)] / (50.2 + 80.5) = 0.986 M