Video: Calculating the Average Rate of Reaction of Hydrochloric Acid with Calcium Carbonate

When excess hydrochloric acid reacts with calcium carbonate, carbon dioxide is produced. If 5 grams of calcium carbonate is consumed in 3 minutes and 20 seconds, what is the average rate of reaction?

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Video Transcript

When excess hydrochloric acid reacts with calcium carbonate, carbon dioxide is produced. If five grams of calcium carbonate is consumed in three minutes and 20 seconds, what is the average rate of reaction?

Hydrochloric acid, HCl, is obviously an acid. And we can see that calcium carbonate, CaCO3, is a carbonate. What might spring to mind is the classic reaction between an acid and a carbonate producing a salt, carbon dioxide, and water. And indeed, the question tells us that carbon dioxide is one of the products. It’s not necessary to answer the question, but I’m going to draw out the chemical equation. Here, we have HCl reacting with solid calcium carbonate. I’ve assumed it’s solid because calcium carbonate isn’t soluble in water. Our products are the soluble salt, calcium chloride; the gas, CO2; and H2O in its liquid form. And we’ll need two HCl in order to balance it.

The question also tells us that we’re consuming five grams of calcium carbonate in three minutes and 20 seconds. Our job is to use this information to calculate the average rate of reaction. When we’re dealing with masses, the average rate of our reaction is equal to the change in mass divided by the time taken. But we’ve been given the time taken in a mixture of units, minutes and seconds. So we should convert this all to minutes or all to seconds. I’m going to do it all to seconds.

There are 60 seconds in one minute, so our time taken is equal to three minutes multiplied by 60 seconds per minute add the 20 seconds which is equal to 180 seconds plus 20 seconds, which is 200 seconds. So our average rate with respect to the calcium carbonate is five grams divided by 200 seconds, which works out as 0.025 grams of calcium carbonate per second.

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