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Balance the following equation using the atom inventory method:

____Cr(MnO4)3 + ____Sr(OH)2 → ____Cr(OH)3 + ____Sr(MnO4)2

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Answer :

The balanced equation is:

Cr(MnO4)3 + 3Sr(OH)2 → 5Cr(OH)3 + 2Sr(MnO4)2

Balancing chemical equation.

To balance this equation using the atom inventory method, we need to make sure that the number of each type of atom is the same on both sides of the equation.

Let's start by looking at the left-hand side of the equation:

Cr(MnO4)3 + Sr(OH)2

We can see that there is one chromium (Cr) atom, three manganese (Mn) atoms, and 12 oxygen (O) atoms on the left-hand side.

Now let's look at the right-hand side of the equation:

Cr(OH)3 + Sr(MnO4)2

Here we have one chromium (Cr) atom, two manganese (Mn) atoms, and 14 oxygen (O) atoms.

So, we need to balance the equation by adding coefficients (numbers in front of each chemical formula) to make sure the number of each type of atom is the same on both sides.

Let's start by balancing the chromium (Cr) atoms:

Cr(MnO4)3 + 3Sr(OH)2 → Cr(OH)3 + Sr(MnO4)2

Now we have one chromium atom on both sides.

Next, let's balance the manganese (Mn) atoms:

Cr(MnO4)3 + 3Sr(OH)2 → Cr(OH)3 + Sr(MnO4)2

We have three Mn atoms on the left-hand side, but only two on the right-hand side. To balance this, we need to add a coefficient of 2 in front of Sr(MnO4)2:

Cr(MnO4)3 + 3Sr(OH)2 → Cr(OH)3 + 2Sr(MnO4)2

Now we have two Mn atoms on both sides.

Finally, let's balance the oxygen (O) atoms:

Cr(MnO4)3 + 3Sr(OH)2 → Cr(OH)3 + 2Sr(MnO4)2

We have 12 O atoms on the left-hand side, but 20 O atoms on the right-hand side. To balance this, we need to add a coefficient of 5 in front of Cr(OH)3:

Cr(MnO4)3 + 3Sr(OH)2 → 5Cr(OH)3 + 2Sr(MnO4)2

Now we have 12 O atoms on both sides.

So the balanced equation is:

Cr(MnO4)3 + 3Sr(OH)2 → 5Cr(OH)3 + 2Sr(MnO4)2

Therefore, the coefficients are:

Cr(MnO4)3 + 3Sr(OH)2 → 5Cr(OH)3 + 2Sr(MnO4)2.

Learn more on balancing chemical equation here https://brainly.com/question/11904811

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