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oee [108]
4 years ago
12

The chemical formula for one repeat unit of the PP polymer is CaHe. Produce a balanced chemical equation for the complete combus

tion of one or more repeat units of PP in oxygen to form carbon dioxide and water (you may find it helpful to show the steps in your working Explain how you can test that your equation is balanced
Chemistry
1 answer:
Vikki [24]4 years ago
7 0

Answer:

2C₃H₆ + 9O₂ ⟶ 6CO₂ + 6H₂O

Explanation:

The repeating unit in PP is C₃H₆.

The unbalanced equation for the combustion of one repeating unit is

C₃H₆ + O₂ ⟶ CO₂ + H₂O

Here's how to balance the equation.

1. Pick the most complicated-looking formula (C₃H₆).

Put a 1 in front of it.

1C₃H₆ + O₂ ⟶ CO₂ + H₂O

2. Balance C.

We have fixed 3C on the left. We need 3C on the right.

Put a 3 in front of CO₂.

1C₃H₆ + O₂ ⟶ 3CO₂ + H₂O

3. Balance H.

We have fixed 6H on the left. We need 6H on the right.

Put a 3 in front of H₂O.

1C₃H₆ + O₂ ⟶ 3CO₂ + 3H₂O

4. Balance O

We have fixed 9O on the right. we need 9O on the left.

Uh, oh. Fractions. we need 4½ O₂.

Multiply every coefficient by two.

2C₃H₆ + O₂ ⟶ 6CO₂ + 6H₂O

Now, we have 18O on the right. Balance O by putting a 9 in front of O₂.

2C₃H₆ + 9O₂ ⟶ 6CO₂ + 6H₂O

All species have a coefficient. The equation should now be balanced.

5. Check that all atoms are balanced

The balanced equation should have the same number of atoms on each side of the reaction arrow.

\begin{array}{ccc}\textbf{Atom} & \textbf{On the left} & \textbf{On the right}\\\text{C} & 6 & 6\\\text{H} & 12 & 12\\\text{O} & 18 & 18\\\end{array}

The balanced equation is

2C₃H₆ + 9O₂ ⟶ 6CO₂ + 6H₂O

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\large \boxed{1. \text{ 0.17 g/L; 2. 3.52; 3. Cl; 4. (a) +3; (b) +4; (c) +6}}

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\rm \stackrel{\hbox{0}}{\hbox{Zn}} + \stackrel{\hbox{0}}{\hbox{ Cl}_{2} }\longrightarrow \stackrel{\hbox{+2}}{\hbox{Zn}}\stackrel{\hbox{-1}}{\hbox{Cl}_{2}}

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Cl has been reduced, so Cl is the oxidizing agent.

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\stackrel{\hbox{$\mathbf{+3}$}}{\hbox{Al}_{2}}\stackrel{\hbox{-2}}{\hbox{O}_{3}}

1O = -2; 3O = -6; 2Al  = +6; 1Al = +3

(b) XeF₄

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