Balancing Reaction Pt 2

Question

Balance the following equations, if no coefficient is needed enter the number 1:

___HClO₄ + ___P₄O₁₀  ---> ___H₃PO₄ + ___Cl₂O₇

Answer 1

Let's start by breaking down each element in our equation into individual molecules and atoms.

___HClO₄ + ___P₄O₁₀  ---> ___H₃PO₄ + ___Cl₂O₇

Our goal is to separate all the molecules into the individual elements -- then count the number of each element.

Left

H - 1

Cl - 1

O - 4 + 10 = 14

P - 4

Right

H - 3

Cl - 2

O - 4 + 7 = 11

P - 1

Now that we've separated our molecules into individual atoms, we compare coefficients. We can start with Chlorine, Cl. There's 1 on the left side of the equation, and 2 on the right side. So to make those numbers match, we multiply the left side of the equation by two.

2 HClO₄ + ___P₄O₁₀  ---> ___H₃PO₄ + ___Cl₂O₇

Let's recount these new coefficients. We have 4 P atoms on the left, and 1 P atom on the right. To make these match, we must multiply the right coefficient by 4.

2 HClO₄ + ___P₄O₁₀  ---> 4 H₃PO₄ + ___Cl₂O₇

We now have 12 H atoms on the right, and just 2 on the left. To make these match, we'll multiply the left by 6.

12 HClO₄ + ___P₄O₁₀  ---> 4 H₃PO₄ + ___Cl₂O₇

Now there are 12 Cl atoms on the left, and 2 Cl atoms on the right. Let's multiply the right by 6.

12 HClO₄ + ___P₄O₁₀  ---> 4 H₃PO₄ + 6 Cl₂O₇

Now let's count up our molecules on both sides again. After counting, if a number doesn't match, add a coefficient.

12 HClO₄ + ___P₄O₁₀  ---> 4 H₃PO₄ + 6 Cl₂O₇

Left

H - 12

Cl - 12

O - 48 + 10 = 58

P - 4

Right

H - 12

P - 4

O - 16 + 42 = 58

Cl - 12

Wonderful! Everything matches. So our final equation is going to be...

12 HClO₄ + P₄O₁₀  ---> 4 H₃PO₄ + 6 Cl₂O₇

Now, you just have to double check that the molecule count matches on both sides, and you're done! Great work!