Stoichiometry/Limiting Reagents

[wvlaxxer81]

Could someone please check my work on some of these problems so I can fix whatever I need to. I need to know if I'm doing this right because our test is tomorrow on this unit.

Final Answers are in BOLD


STOICHIOMETRY

1. 150 grams of potassium chlorate is decomposed

A. Balanced equation = 2KClO3 --> 2KCl + 3O2

B. How many moles of potassium chlorate are there? 150g KClO3 x 1 mol / 122.5g KClO3 = 1.22 mol KClO3

C. How many moles of potassium chlorate are formed? 1.22 mol KClO3 x 2 KCl / 2KClO3 = 1.22 mol KCl

D. How many moles of oxygen are formed? 1.22 mol KClO3 x 3 O2 / 2KClO3 = 1.83 mol O2

E. What is the mass of the potassium chloride? 74.5g x 1.22 mol = 90.89g KCl

F. What is the mass of the oxygen? 32g x 1.83 mol = 58.56g O2

G. What is the volume of the oxygen? 1.83 mol O2 x 22.4L / 1 mol = 40.992 L O2


2. Iron (III) reacts with hydrochloric acid. If 25 liters of hydrogen gas are formed at STP.

A. Balanced Equation. 2Fe3 + 18 HCl --> 6FeCl3 + 9H2

B. What is the mass of the Iron (III)? 25/22.4 = 1.12 mol H2

1.12 mol H2 x 2 mol Fe3/9 mol H2 = 0.25 mol Fe3

0.25 mol Fe3 x 56g Fe3 = 14g Fe3

C. What is the mass of the hydrochloric acid? 1.12 mol H2 x 18 mol HCl/9 mol H2 = 2.24 mol HCl

2.24 mol HCl x 36.5g HCl/1 mol HCl = 81.76g HCl

D. What is the mass of the Iron (III) Chloride? 1.12 mol H2 x 6 mol FeCl3/9 mol H2 = 0.75 mol FeCl3

0.75 mol x 91.5g FeCl3/1 mol FeCl3 = 68.63g FeCl3

E. What is the mass of the Hydrogen gas? 1.12 mol x 2g H2/1 mol H2 = 2.24g H2


3. Phosphorous reacts with Oxygen to yield Diphosphorous Pentaoxide (gas). If 50 liters of Oxygen is used:

A. Balanced Equation. 4P +5O2 --> 2(P2O5)

B. What is the volume of the Diphosphorous Pentaoxide? 50 L/22.4L = 2.23 mol O2

2.23 mol O2 x 2 mol P2O5/5 mol O2 = 0.89 mol P2O5

0.89 mol P2O5 x 22.4 L/1 mol P2O5 = 19.94 L P2O5

C. What is the mass of the Diphosphorous Pentaoxide? 0.89 mol P2O5 x 142g P2O5/1 mol P2O5 = 126.38g P2O5

D. What is the mass of the Phosphorous? 2.23 mol O2 x 4 mol P/5 mol O2 = 1.78 mol P

1.78 mol P x 31g P/1 mol P = 55.18g P

E. What is the mass of the Oxygen? 2.23 mol O2 x 32g O2/1 mol O2 = 71.36g O2

F. How many molecules of oxygen was used? 2.23 mol O2 x 6.02x10^23 mc O2 = 13.42x10^23 mc O2

G. How many molecules of Diphosphorous Pentaoxide were formed? 0.89 mol P2O5 x 6.02x10^23 mc P2O5 = 5.36x10^23 mc P2O5

H. How many atoms of Phosphorous reacted? 1.78 mol P x 6.02x10^23 atoms P = 10.72x10^23 at P



LIMITING REAGENTS


1. 2.8 moles of H2O reacts with 3.1 moles of Cl2O7

A. Balanced equation. Cl2O7 + H2O --> 2HClO4

B. What is the limiting reagent? H2O

C. How many moles of the product are produced? 2.8 mol H2O x 2 mol HClO4/1 mol H2O = 5.6 mol HClO4

D. How many moles of the excess reagent remains unreacted? 3.1 mol Cl2O7 - 2.8 mol H2O = 0.3 mol Cl2O7


2. 9.8 moles of Al reacts with 12.4 moles of CuSO4

A. Balanced equation. 2Al + 3CuSO4 --> 3Cu + Al2(SO4)3

B. What is the limiting reagent? Aluminum

C. How many moles of Al2(SO4)3 are produced? 9.8 mol Al x 1 mol Al2(SO4)3/2 mol Al = 4.9 mol Al2(SO4)3

D. How many moles of the excess reagent remain? 12.4 mol CuSO4 - 9.8 mol Al = 2.6 mol CuSO4


3. 2.4 grams of Na reacts with 36 grams of H2O

A. Balanced Equation. 2Na + H2O --> Na2O + H2

B. What is the limiting reagent? H2O

C. How many moles of H2 are formed? 36g H2O x 1 mol H2O/18g H2O = 2 mol H2O

2 mol H2O x 1 mol H2/1 mol H2O = 2 mol H2

D. How many grams of H2 are formed? 2 mol H2 x 2g H2/1 mol H2 = 4g H2

E. How many grams of the excess reagent are left? 2.4g Na - 1.2g = 1.2g Na

F. How many liters of H2 are formed? 2 mol H2 x 22.4 L H2/1 mole H2 = 44.8 L H2


4. 123g of Fe reacts with 56g of O2

A. Balanced Equation. 4Fe + 3O2 --> 2Fe2O3

B. What is the limiting reagent? O2

C. How many grams of Fe2O3 are formed? 56g O2 x 2 mol Fe2O3/3 mol O2 = 37.3g Fe2O3

D. How many grams of the excess reagent are left? 123g Fe x 3 mol O2/4 mol Fe = 92.25g Fe

123g Fe - 92.25g Fe = 30.75g Fe


All help is greatly appreciated. Thank you.

 

[Borek]

1. 2.8 moles of H2O reacts with 3.1 moles of Cl2O7

A. Balanced equation. Cl2O7 + H2O --> 2HClO4

B. What is the limiting reagent? H2O

C. How many moles of the product are produced? 2.8 mol H2O x 2 mol HClO4/1 mol H2O = 5.6 mol HClO4

D. How many moles of the excess reagent remains unreacted? 3.1 mol Cl2O7 - 2.8 mol H2O = 0.3 mol Cl2O7

Correct.

Now you know answers to two questions ;)

 

[Blueshift5]

Your work looks correct to me! Just make sure to double check your calculations and units to ensure accuracy. Good luck on your test tomorrow!