# If 730 mL of 0.17808 M aqueous HClO4 reacts stoichiometricallyaccording to the balanced equation how many...

If 730 mL of 0.17808 M aqueous HClO4 reacts stoichiometrically

according to the balanced equation how many milliliters of 4.08 M aqueous Ba(ClO4)2 are produced?

Ba(OH)2(*aq*) + 2HClO4(*aq*) → Ba(ClO4)2(*aq*) + 2H2O(*l*)

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To start, we need to get what we know in terms of moles. If we do this, then we are able to use the chemical equation with its associated molar ratios to solve the problem.

To calculate the number of moles given molarity and volume, we simply multiply M by liters because M is in terms of moles/liter. Therefore, we need to convert the given volume of HClO4 to liters of HClO4:

Liters HClO4 = 730 mL HClO4 * 1L / 1000 mL = .73 L HClO4

Here you will notice that mL cancels out as a unit and only L remains. Now, we can calculate the moles of HClO4. We will first convert molarity (M) to moles/liter:

Moles HClO4 = 0.73 L HClO4 * 0.17808 mol HClO4 / 1 L HClO4

Here, L HClO4 cancels out and we are left with the following result. Remember, we have only 2 significant figures, so we need to round!

Moles HClO4 = 0.13 mol HClO4

Now we can use the chemical equation. Notice that in the chemical equation that for every 2 moles of HClO4 used, 1 mole of Ba(ClO4)2 is produced. Therefore, the number of moles of Ba(ClO4)2 created in the reaction will be half of the number of moles of HClO4 we put in!

Moles Ba(ClO4)2 = 0.13 mol HClO4 * 1 mol Ba(ClO4)2 / 2 mol HClO4

Moles Ba(ClO4)2 = 0.065 mol Ba(ClO4)2

Again, we keep to 2 significant figures, which is moot for this particular equation.

To find the number of mL Ba(ClO4)2(*aq*) produced at the given molarity, we do the opposite of what we did above to find the moles of HClO4. We will divide our moles of Ba(ClO4)2 by the given molarity. Again, we will convert M to mol/L:

Volume Ba(ClO4)2 = 0.065 mol Ba(ClO4)2 / (4.08 mol Ba(ClO4)2 / 1 L Ba(ClO4)2)

Because we are dividing by a quotient, we flip the fraction and multiply. Notice that mol Ba(ClO4)2 cancels out, leaving us with the following relation:

Volume Ba(ClO4)2 = 0.065/4.08 L Ba(ClO4)2 = 0.016 L Ba(ClO4)2

The problem asks for the answer in mL, so we must convert by multiplying by 1000:

Volume Ba(ClO4)2 = 0.016 L Ba(ClO4)2 * 1000 mL/ 1 L

**Volume Ba(ClO4)2 = 16 mL Ba(ClO4)2**

Notice that problems like this sum up to simply multiplying and dividing in a way that cancels out units! Nothing more, nothing less.

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