In: Chemistry
If 23 g. of Ferrous sulfide is added to 30 g. of oxygen and 6 g. of ferric oxide is produced in the reaction of problem # 38 . The limiting reagent is
a. Ferrous sulfide b. oxygen c. Ferric oxide d. sulfur dioxide
If 23 g. of Ferrous sulfide is added to 30 g. of oxygen and 6 g. of ferric oxide is produced in the reaction of problem # 38 . The excess reagent is
a. Ferrous sulfide b. oxygen c. Ferric oxide d. sulfur dioxide
1. The number of mole(s) of ferric oxide in 30 g. of ferrous sulfide in problem # 38.
a. 0.03 b. 0.13 c. 0.26 d. 0.33
2. The amount of excess reagent in grams in problem # 38.
a. 22.88 b. 20.45 c. 15.44 d. 6.00
3. Theoretical yield in grams in problem # 38.
a. 22.88 b. 20.45 c. 15.44 d. 6.00
4. The actual yield in grams in problem #38.
a. 22.88 b. 20.45 c. 15.44 d. 6.00
5. Percent yield in problem #38.
a. 16 b.26 c.36 d.46
6. A sample of gas occupies a volume of 225 ml. at a pressure of 720 torr and a temperature of 20 o C. Calculate the new pressure in torr, if the volume is increased to 350 ml, at constant temperature.
a. 263 b. 363 c. 463 d. 563
7. A sample of gas occupies a volume of 275 ml at 20 oC and 1 atm pressure. Calculate the volume in ml. of the gas at 0oC and 1 atm pressure.
a. 256 b. 356 c. 456 d. 556
8. Calculate the total pressure, in torr, of a mixture of gases, each of which exerts the given partial pressure: H2, 150 torr ; N2, 210 torr; He, 320 torr.
a. 580 b 680 c. 780 d. 880
9. A sample of gas occupies a volume of 512 ml at 20 oC and 740 torr. What volume in ml. would this gas occupy at STP.
a. 364 b. 464 c. 564 d. 664
10. A sample of oxygen gas collected over water occupies a volume of 210 ml at 22 oC and 750 torr pressure. Calculate the volume in ml of dry oxygen at STP.
a. 487 b. 387 c. 287 d. 187
balanced equation :
4 FeS + 7O2 -------------> 2 Fe2O3 + 4 SO2
351.64 g 224 g 319.38 g 256.24 g
23 g 30 g ??
here limting reagent if FeS
excess reagent is b) oxygen
1. The number of mole(s) of ferric oxide in 30 g. of ferrous sulfide in problem # 38.
moles of FeS = 30 / 87.91 = 0.341
4 mol FeS -----------> 2 mol Fe2O3
0.341 mol FeS ----------> ??
moles of Fe2O3 = 0.17
2.)
amount of excess reagent = 30 - (14.65) = 15.4 g
answer : c. 15.44 g
3.)
theroretical yield = 23 x 319.38 / 351.64 = 20.88 g
answer : 20.88 g
4.)
option d) 6.00 g
5)
% yield = 6 / 20.88 ) x 100
= 26 %