## Methodology

Problem One
Phosphorus trichloride (PCl3) reacts with water to form HCl and H3PO3. Write an equation for this reaction.

Step #1. Identify the reactants and products.

reactants: PCl3 and H2O
products: HCl and H3PO3

Step #2. Write a temporary, unbalanced "equation".

PCl3 + H2O → HCl + H3PO3

Step #3. Count the number of phosphorus atoms on both sides of the arrow.

One P on the left (in PCl3), one P on the right (in H3PO3)

Step #4. Count the number of oxygen atoms on both sides.

One oxygen on left, three on the right.

Step #5. Make the number of oxygen atoms equal by multiplying H2O by 3.

PCl3 + 3 H2O → HCl + H3PO3

Step #6. Now count up the hydrogens in this new equation.

6 hydrogens on left, four on right.

Step #7. Equalize the hydrogens by multiplying HCl by 3 (if you tried to change H3PO3 you would then affect the number of P atoms, which in turn would affect the others.)

PCl3 + 3 H2O → 3 HCl + H3PO3

Step #8. Count the number of chlorine atoms on both sides to be sure that all elements are balanced.

three on left, three on right

Problem Two
Ethene (C2H4) reacts with oxygen to form CO2 and water. Write an equation for this reaction.

Step #1. Identify the reactants and products.

reactants: C2H4 and O2
products: CO2 and H2O
Notice that you must know that elemental oxygen exists as O2 molecules.

Step #2. Write a temporary, unbalanced "equation".

C2H4 + O2 → CO2 + H2O

Step #3. Count the number of carbon atoms on both sides of the arrow.

two on the left, one on the right

Step #4. Make the number of carbon on the right equal to the number on the left by multiplying CO2 by 2.

C2H4 + O2 → 2 CO2 + H2O

Step #5. Make the number of hydrogens on both sides equal.

C2H4 + O2 → 2 CO2 + 2 H2O

Step #6. Make the number of oxygen atoms equal by using an appropriate coefficient for O2.

C2H4 + 3 O2 → 2 CO2 + 2 H2O

Step #7. Check to be certain that you have whole number coefficients and that all elements are balanced.

Problem Three
In the reaction of PCl3 with water:
PCl3 + 3 H2O → 3 HCl + H3PO3
determine how many moles of PCl3 would be required to form 0.2 mole HCl.

Step #1. Make certain that your "equation" is balanced.

It is. We did it in problem 1.

Step #2. Examine the coefficients of the PCl3 and HCl and verbalize the relationship.

One mole of PCl3 will be consumed to form three moles of HCl.

Step #3. Determine how many moles of PCl3 would be consumed in making one mole of HCl. You may want to divide all coefficients by 3 in order to clearly see the relationship (which we already expressed in step#2).

1/3 PCl3 c+ H2O → HCl + c1/3 H3PO3
Thus, 1/3 of a mole of PCl3 will be consumed in making one mole of HCl.

Step #4. Now apply your mole-mole relationship to the formation of 0.2 mole HCl.

If one-third mole of PCl3 is required to make one mole of HCl, one-third of 0.2 mole will be required to make 0.2 mole. That is,
0.20 mole x 1/3 = 0.067 mole PCl3

Step #5. Redo step#4 using dimensional analysis.

0.20 mole HCl x (1 mole PCl3/ 3 mole HCl) = 0.067 mole PCl3

Step #6. Redo step#4 using proportions.

One mole HCl is to one-third mole PCl3 as 0.2 mole HCl is to x mole PCl3.
(1 mole HCl / 0.33 mole PCl3) = (0.20 mole HCl / x mole PCl3)
x = 0.067 mole PCl3

Problem Four
In the reaction of Fe2O3 with CO
Fe2O3 + 3 CO → 2 Fe + 3 CO2
calculate the number of moles of CO required to form 0.5 mole Fe.

Step #1. Make certain that your "equation" is balanced.

It is.

Step #2. Examine and verbalize the mole-mole relationship between CO and Fe.

Three moles of CO are required to form 2 moles of Fe.

Step #3. Apply the mole-mole relationship between CO and Fe to the formation of 0.5 mole Fe.

0.50 mole Fe x (3 moles CO/ 2 moles Fe) = 0.75 mole CO
Notice that the coefficients tell you that the number of moles of CO must be greater than the number of moles of Fe. Also notice that the moles of iron cancel leaving you with units of moles CO.

Problem Five
In the reaction of Fe2O3 with CO
Fe2O3 + 3 CO → 2 Fe + 3 CO2
calculate the number of grams of Fe formed from the complete consumption of 3.4 g CO.

Step #1. Make certain that the "balanced" equation is correct.

It is.

Step #2. Examine the mole-mole relationship between CO and Fe and verbalize it.

The consumption of 3 moles of CO will produce 2 moles of Fe, or two-thirds of a mole of Fe are produced for every mole of CO consumed.

Step #3. Calculate the number of moles of CO consumed.

(3.4 g CO / 34 g per mole) = 0.10 mole CO

Step #4. Calculate the number of moles of Fe produced.

0.10 mole CO x (2 moles Fe / 3 moles CO) = 0.067 mole Fe

Step #5. Convert moles of Fe to mass of Fe.

0.067 mole Fe x 55.8 g/mol = 3.7 g Fe

Problem Six
A 12 g sample of tin is heated with a 25 g sample of I2. What is the maximum amount of SnI4 that can be obtained?

Step #1. Write an equation for the reaction.

Sn + 2 I2 → SnI4

Step #2. Calculate the mole amounts of Sn and I2.

Sn: 12 g / 119 g/mole = 0.10 mole
I2: 25 g / 254 g/mole) = 0.10 mole

Step #3. Determine which reagent is the limiting reagent by applying the 1 to 2 mole-mole relationship between tin and I2.

We have 0.10 mole tin and need twice that amount of I2 to react completly with it. We do not have twice that amount. Therefore the tin is in excess and the I2 is the limiting reagent.

Step #4. Calculate the number of moles of SnI4 obtained from the limiting reagent.

0.10 mole I2 x (1 mole SnI4 / 2 mole I2) = 0.050 mole SnI4.

Step #5. Convert moles of product to mass of product.

0.050 mole SnI4 x 626.3 g/mol = 31 g SnI4

Problem Seven
Consider the following reaction:
4 HCl + MnO2 → Cl2 + MnCl2 + 2 H2O
If 20 mL of 1 M HCl is added to 2.0 g MnO2 and the reaction has an 85% yield, how many grams of MnCl2 are formed?

Step #1. How many moles of HCl are there in 20 mL of 1 M HCl?

0.020 L x 1 mole/L = 0.020 moles of HCl

Step #2. How many moles of MnO2 are present?

2.0 g / 87 g/mole = 0.023 mole

Step #3. Which reagent is the limiting reagent?

The 0.023 mole MnO2 requires four times that mole amount (0.092 mole) of HCl. This amount is not present. Therefore, the HCl is the limiting reagent.

Step #4. What is the relationship between the number of moles of HCl and the number of moles of MnCl2?

For every mole of HCl consumed, one-fourth of a mole of MnCl2 is formed.

Step #5. How many moles of MnCl2 will be formed from 0.020 moles of HCl if the reaction goes to completion?

0.020 moles HCl x (1 mole MnCl2/ 4 moles HCl) = 0.0050 mole MnCl2

Step #6. How many moles of MnCl2 will be formed from 0.020 moles of HCl with an 85% yield?