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Chapter 2: Problem 12

In one experiment, the reaction of \(1.00 \mathrm{g}\) mercury and an excess of sulfur yielded \(1.16 \mathrm{g}\) of a sulfide of mercury as the sole product. In a second experiment, the same sulfide was produced in the reaction of \(1.50 \mathrm{g}\) mercury and \(1.00 \mathrm{g}\) sulfur. (a) What mass of the sulfide of mercury was produced in the second experiment? (b) What mass of which element (mercury or sulfur) remained unreacted in the second experiment?

Short Answer

Expert verified
The mass of the sulfide of mercury produced in the second experiment was 1.74g and the mass of sulfur that remained unreacted was 0.76g.

Step by step solution

01

Calculate Mass of Sulfide Produced in Second Experiment

The amount of mercury is 1.5 g in the second experiment. As the sulfide of mercury is produced by the reaction of 1g of mercury, we can calculate the mass of the sulfide produced in the second experiment by setting up a proportion: \(1.00 g \, mercury = 1.16 g \, sulfide \) and \(1.50 g \, mercury = x g \, sulfide\), thus solving for x gives \(x = 1.5 * \frac{1.16}{1} = 1.74 g \,of\, sulfide\).
02

Calculate Mass of Reacted Sulfur

From the first experiment, we know that the mass of sulfur which reacted with 1g of mercury is \(1.16 g - 1 g = 0.16 g\). Thus, the mass of the sulfur reacted with 1.5g of mercury in the second experiment can again be calculated using a proportion. This gives: \(1 g \, mercury = 0.16 g \, sulfur\) and \(1.5 g \, mercury= y g\, sulfur \), thus solving for y gives \(y = 1.5 * \frac{0.16}{1} = 0.24 g \,of \, reacted \,sulfur\).
03

Determine Element Remaining Unreacted

In the second experiment, we started with 1 g of sulfur. The mass of reacted sulfur was calculated earlier to be 0.24 g, so the amount of sulfur remaining is \(1 g - 0.24 g = 0.76 g\). As there is still sulfur left, but no mercury, sulfur is the element that remained unreacted.

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