In the following reaction, the rate of sulfur formation was determined to be 0.002 grams per second. Na2S2O3 gas plus two HCl gas reacts to form two NaCl aqueous plus H2O liquid plus SO2 gas plus S solid. How much sulfur was formed after five minutes?
In this reaction, sodium thiosulfate reacts with hydrochloric acid to form sodium chloride, water, sulfur dioxide gas, and sulfur. We need to determine how much sulfur is formed in this reaction. And we’re given the rate of sulfur formation. This is the rate of this chemical reaction. Looking at the units, we can tell this rate is by mass. The rate of reaction can be defined in terms of the formation of a product or in terms of the consumption of a reactant. The Δ symbol means change in. So, for both equations, the rate of reaction is the change in mass divided by the change in time.
You’ll notice that the rate of reaction for the reactants is defined as a negative quantity. This problem wants us to calculate how much sulfur was formed, which means we’ll be calculating a mass. Since sulfur is a product in the reaction, we’ll use the rate equation for the formation of our product. We can create an expression to calculate the mass by multiplying both sides of the rate equation by the change in time. Now we can plug in the values from the problem.
The rate is 0.002 grams per second, and the time is five minutes. We’ll need our units of time here to cancel so that we can end up in units of grams. But right now, they don’t match. So, let’s convert our minutes into seconds. There are 60 seconds in one minute. So, we can convert from minutes to seconds by multiplying by 60, which gives us 300 seconds. Now, the units cancel. Multiplying 0.002 by 300 gives us 0.6 grams. So, 0.6 grams of sulfur formed after five minutes.