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Problem 73

Calculate the mass (g) of each product formed when 174 $\mathrm{g}$ of silver sulfide reacts with excess hydrochloric acid:

$$

\mathrm{Ag}_{2} \mathrm{S}(s)+\mathrm{HCl}(a q) \longrightarrow \mathrm{AgCl}(s)+\mathrm{H}_{2} \mathrm{S}(g)[\text { unbalanced }]

$$

Answer

$\mathrm{m}=\mathrm{n}(\mathrm{AgCl})^{*} \mathrm{Mr}(\mathrm{AgCl})=201,22 \mathrm{g}$

$\mathrm{m}\left(\mathrm{H}_{2} \mathrm{S}\right)=\mathrm{n}^{*} \mathrm{Mr}\left(\mathrm{H}_{2} \mathrm{S}\right)=0,702^{*}(2+32,06)=23,91 \mathrm{g}$

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## Discussion

## Video Transcript

So we're keeping the reaction between ah sue for selfies and also hydrochloric acid. And we have to find out the mass of each of the part that they're super Karan also. Ah, us h two s, the Hodgins survive. All right, so we're giving this equation base and know that bonds. So for silver rehab, £2 toe Camco Yashin. So Ah, we sure we have the same number? Oh, seal first, Which us, too, is over. And then we also have make sure we have the same number off. Ah ah, Call Corrine. So we have tea. I put two in Illinois Excel. Okay, so we have to exhale. And then we have two hydrogen and also church or right. And also we have oneself for 11. Oneself are the right so they can congressional sponsors. That looks stab is we're going to find the lumber most for our suit yourself. Full for silver has won those 7.9. Ah, for the move masks. So we have to off them and then we're going thio plus Ah, sulfur. So it will be, Ah to 47.87 Grampa herbal for the mess. So we're going to take our master bi bi minimus, we would have syrup on +70 it's you. Okay. So again, we're looking at the more the way show. So the more the ratio between super self I and also Cervical writes 1 to 2. It's 1 to 2. So where we have one off it, we have to use a vehicle. Right? So we have cereal point saving odes 702 Both, uh, number most we can hope about you. And then we're 1.44 uh, most for civic. All right. And then for ah 100 himself I the 100 flow vice that I can see that a mile away. So it's 1 to 1. So they should have the same number almost the same time. Bumbles. You know, the father mass, which is that you find out Ah, Mother Mass for off a supercar in the Mumbai By the limbo most So we have the album we have Super. It's one those 7.9 and they were one call right? 45 or 45 And then we promoted by everything together you should be able to find as to a one, um Graham for the full right and then for hydrogen sulfur. So for the Dodgers, all Celso survive. We have soup on 702 times. Ah, the mullahs mass to hydrogen and also with one's over. So for and then you should be able to find this off there on 23.9 crab and those are the

## Recommended Questions

Consider the unbalanced chemical equation

$$\mathrm{Ag}(s)+\mathrm{H}_{2} \mathrm{S}(g) \rightarrow \mathrm{Ag}_{2} \mathrm{S}(s)+\mathrm{H}_{2}(g)$$

Balance the equation. Identify the mole ratios that you would use to calculate the number of moles of each product that would form for a given number of moles of silver reacting.

Silver chloride, used in silver plating, contains $75.27 \%$ Ag. Calculate the mass of silver chloride in grams required to make $4.8 \mathrm{g}$ of silver plating.

The formation of water-insoluble silver chloride is useful in the analysis of chloride-containing substances. Consider the following unbalanced equation: $\mathrm{BaCl}_{2}(\mathrm{aq})+\mathrm{AgNO}_{3}(\mathrm{aq}) \longrightarrow \mathrm{AgCl}(\mathrm{s})+\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})$

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$$

\mathrm{Al}_{2} \mathrm{S}_{3}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{Al}(\mathrm{OH})_{3}+\mathrm{H}_{2} \mathrm{S}[\text { unbalanced }]

$$

How many grams of the excess reactant remain?

Silver sulfide, the tarnish on silverware, comes from reac- tion of silver metal with hydrogen sulfide $\left(\mathrm{H}_{2} \mathrm{S}\right)$ :

$$\mathrm{Ag}+\mathrm{H}_{2} \mathrm{S}+\mathrm{O}_{2} \longrightarrow \mathrm{Ag}_{2} \mathrm{S}+\mathrm{H}_{2} \mathrm{O} \text { Unbalanced }$$

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Given the amounts of reactants shown, calculate the theoretical yield in grams of the product of each of these unbalanced chemical equations.

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$30.0 \mathrm{g} \quad 12.0 \mathrm{g} \quad$ ? $\mathrm{g}$

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$24.3 \mathrm{g} \quad 10.0 \mathrm{g}$

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c. $\mathrm{CuCl}_{2}(a q)+\mathrm{Zn}(s) \rightarrow \mathrm{Cu}(s)+\mathrm{ZnCl}_{2}(a q)$

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Given a quantity in moles of reactant or product, use a mole-mole factor from the balanced chemical equation to calculate the number of moles of another substance in the reaction.

Write all the mole-mole factors for each of the following chemical equations:

a. $2 \mathrm{Al}(s)+3 \mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{AlCl}_{3}(s)$

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For each of the following unbalanced equations, calculate how many moles of the second reactant would be required to react completely with exactly 25.0 g of the first reactant. Indicate clearly the mole ratio used for each conversion.

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c. NaHSO $_{3}(a q)+\mathrm{NaOH}(a q) \rightarrow$

$\quad \mathrm{Na}_{2} \mathrm{SO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$

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$\quad \mathrm{KCl}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)$

Calculate the mass (g) of each product formed when 43.82 $\mathrm{g}$ of diborane $\left(\mathrm{B}_{2} \mathrm{H}_{6}\right)$ reacts with excess water:

$$

\mathrm{B}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{3} \mathrm{BO}_{3}(s)+\mathrm{H}_{2}(g)[\text { unbalanced }]

$$

Balance each chemical equation.

a. Na2S(aq) + Cu(NO3)2(aq)-NaNO3(aq) + CuS(s)

b. N2H4(l)-NH3(g) + N2(g)

c. HCl(aq) + O2(g)-H2O(l) + Cl2(g)

d. FeS(s) + HCl(aq)-FeCl2(aq) + H2S(g)