Wired Chemist

reactants: PCl₃ and H₂O
products: HCl and H₃PO₃

Step #2. Write a temporary, unbalanced "equation".
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PCl₃ + H₂O → HCl + H₃PO₃

Step #3. Count the number of phosphorus atoms on both sides of the arrow.
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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.
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One oxygen on left, three on the right.

Step #5. Make the number of oxygen atoms equal by multiplying H₂O by 3.
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PCl₃ + 3 H₂O → HCl + H₃PO₃

Step #6. Now count up the hydrogens in this new equation.
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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.)
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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.
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three on left, three on right

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".
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C₂H₄ + O₂ → CO₂ + H₂O

Step #3. Count the number of carbon atoms on both sides of the arrow.
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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.
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C₂H₄ + O₂ → 2 CO₂ + H₂O

Step #5. Make the number of hydrogens on both sides equal.
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C₂H₄ + O₂ → 2 CO₂ + 2 H₂O

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

It is. We did it in problem 1.

Step #2. Examine the coefficients of the PCl₃ and HCl and verbalize the relationship.
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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).
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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.
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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.
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0.20 mole HCl x (1 mole PCl₃/ 3 mole HCl) = 0.067 mole PCl₃

Step #6. Redo step#4 using proportions.
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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

It is.

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

It is.

Step #2. Examine the mole-mole relationship between CO and Fe and verbalize it.
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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.
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(3.4 g CO / 34 g per mole) = 0.10 mole CO

Step #4. Calculate the number of moles of Fe produced.
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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.
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0.067 mole Fe x 55.8 g/mol = 3.7 g Fe

Sn + 2 I₂ → SnI₄

Step #2. Calculate the mole amounts of Sn and I₂.
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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.
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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.
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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.
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0.050 mole SnI₄ x 626.3 g/mol = 31 g SnI₄

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

Step #2. How many moles of MnO₂ are present?
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2.0 g / 87 g/mole = 0.023 mole

Step #3. Which reagent is the limiting reagent?
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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?
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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?
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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?
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0.0050 mole MnCl₂ x 0.85 = 0.0043 mole MnCl₂

Step #7. How many grams of MnCl₂ will be formed?
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0.0042 mole x 126 g/mole = 0.54 g