Silver sulfide, the tarnish on silverware, comes from reac-...

Silver sulfide, the tarnish on silverware, comes from reac- tion of silver metal with hydrogen sulfide $\left(\mathrm{H}_{2} \mathrm{S}\right)$ : $$\mathrm{Ag}+\mathrm{H}_{2} \mathrm{S}+\mathrm{O}_{2} \longrightarrow \mathrm{Ag}_{2} \mathrm{S}+\mathrm{H}_{2} \mathrm{O} \text { Unbalanced }$$ (a) Balance the equation. (b) If the reaction were used intentionally to prepare $\mathrm{Ag}_{2} \mathrm{S}$ , how many grams would be formed from 496 g of Ag, 80.0 $\mathrm{g}$ of $\mathrm{H}_{2} \mathrm{S},$ and 40.0 $\mathrm{g}$ of $\mathrm{O}_{2}$ if the reaction takes place in 95$\%$ yield based on the limiting reactant?

Silver sulfide, the tarnish on silverware, comes from reac- tion of silver metal with hydrogen sulfide $\left(\mathrm{H}_{2} \mathrm{S}\right)$ :
$$\mathrm{Ag}+\mathrm{H}_{2} \mathrm{S}+\mathrm{O}_{2} \longrightarrow \mathrm{Ag}_{2} \mathrm{S}+\mathrm{H}_{2} \mathrm{O} \text { Unbalanced }$$
(a) Balance the equation.
(b) If the reaction were used intentionally to prepare $\mathrm{Ag}_{2} \mathrm{S}$ , how many grams would be formed from 496 g of Ag, 80.0 $\mathrm{g}$ of $\mathrm{H}_{2} \mathrm{S},$ and 40.0 $\mathrm{g}$ of $\mathrm{O}_{2}$ if the reaction takes place in 95$\%$ yield based on the limiting reactant?

Key Concepts

Chemical Equation Balancing

Balancing chemical equations ensures that the same number of atoms of each element is present on both sides of the reaction, adhering to the law of conservation of mass. It involves adjusting coefficients so that the total count of individual atoms remains equal before and after a chemical reaction.

Stoichiometry Calculations

Stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. It uses balanced chemical equations to calculate the amount (in moles or grams) of each substance involved, based on the mole ratios determined by the stoichiometric coefficients.

Limiting Reactant

The limiting reactant is the substance that is completely consumed first in a chemical reaction, thereby determining the maximum amount of product that can be formed. Identifying the limiting reactant is essential for calculating theoretical yields in a reaction.

Percent Yield

Percent yield is a measure of the efficiency of a chemical reaction, calculated by comparing the actual yield obtained from the reaction to the theoretical yield predicted by stoichiometry. It accounts for losses and imperfections in the reaction process, expressed as a percentage.

Transcript

00:01 First, balance the reaction.

00:05 I would like to start by balancing the sulfur atoms, because they seem to be, they only show up once on each side.

00:13 That means that there are one molecule of water, two atoms of silver, and then i get this weird thing where i find that i need half an oxygen atom.

00:26 So i can't really use these coefficients.

00:28 I'm going to go ahead and double them.

00:31 So instead of two, i have four.

00:34 Instead of one there's two one two and two so there's my balanced chemical reaction this problem is a limiting reactant problem because we're given the masses of three of the reactants which seems a lot more complicated than what we're used to with two reactants but the process is the same i'm going to divide each of these by their respective molar masses so for silver that's 107 .87 grams per mole, that's my silver, for h2s, that's 34 .08 grams per mole, and for o2, that's 16 .00 grams per mole.

01:23 And now that i have number of moles of each, this is 4 .60 moles of silver, 2 .35 moles of h2s, and 1 .25 moles of 02.

01:52 I do the comparison of all the different moles, and i notice that this number is more than greater than one half of that number, and this value is greater than one quarter of this value.

02:23 So think of it this way.

02:27 4 .60 moles of silver needs 2 .30 moles of h2s, and i have more than that.

02:42 So that's an excess.

02:45 And this many moles of silver needs divide by 4, 1 .15 moles of 0 .02.

02:57 So again, we have more than that...

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