A 50.00mL solution containing NaBr was treated with excess AgNO_3 precipitate 0.

Question

A 50.00mL solution containing NaBr was treated with excess AgNO_3 precipitate 0.2146g of AgBr (FM 187.772). What is the molarity of NaBr in the solution? B. A 6.50g of petrochemical sample containing C, H, S, N, and O was subjected to a combustion analysis What is the percentage of S and N in the sample if 300 mg of SO_2 and 225 mg NO_2, respectively were produced during the analysis? C. List four examples of implications of high levels of S and N in gasoline D. A 5.22 g of sample containing impure calcium carbonate was treated with an ammonium oxalate and precipitated as calcium oxalate. The precipitate was washed and dried in the oven at elevated temperature of 950 degree C and weighed as calcium oxide. What is the percentage of calcium carbonate in the sample, if the weight of the dried precipitate was 450 mg?

Explanation / Answer

2. A ) AgNO3 + NaBr -------- NaNO3 + AgBr

1 mole 1 mole 1 mole 1 mole

Number of moles of AgBr produced in the procss is = 0.2146 / 187.772 = 0.00114287 moles.

As the reaction is 1:1 so no. of moles for NaBr is also the same

thus the molarity of NaBr = 0.00114287 / 0.05 = 0.0228574 M of NaBr.

2.C. Sulphur and Nitrogen has a great effect on gasoline :-

a) They have effective impactc in emission control system and contribute to air pollution.

b) this content doesnot correlate with the emissions of the critria pollutants from the catalyst free engine.

c) SO2 and NO2 factors are expressed in terms of amount of sulphur and nitrogen in the fuel. It is meant to be assumed that their compounds are captured in the procedure of particulate testing and are included in the particulate emission factor.

d) They mostly cause affects exhaust CO, VOC emission by lowering the catalyst efficiency

2.d CaCO3 + (NH4)2C2O4 ------- CaC2O4 +(NH4)2CO3

CaC2O4.H2O ------ CaC2O4 + H2O

CaC2O4 ------ CaCO3 + CO

CaCO3 ----- CaO + CO2

at a tempararture of 950 0C H2O , CO,CO2 all dissapear

so CaO weigh was 450 mg

so pure CaCO3 present in the sample is = 450/5220 = 0.08620

% of pure calcium carbonate -- = 0.08620 *100 = 806206 %

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