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Ad libitum [116K]
4 years ago
9

Na2S+2AgNO3 = Ag2S + NaNO3 If 2.86g of Ag2S are actually produced by a reaction between an excess of Na2S and 4.27g of AgNO3 the

n what is the percent yield of Ag2S
A. 3.15%
B. 45.9%
C. 61.2%
D. 91.0%
Chemistry
1 answer:
Advocard [28]4 years ago
7 0

Answer:

D. 91.0%

Explanation:

Hello,

In this case, for the given chemical reaction:

Na_2S+2AgNO_3 \rightarrow Ag_2S + 2NaNO_3

Next, since silver nitrate (molar mass 169.87 g/mol) is in a 2:1 molar ratio with silver sulfide (molar mass 247.8 g/mol), we compute its theoretical yield as shown below:

m_{Ag_2S}^{theoretical}=4.27gAgNO_3*\frac{1molAgNO_3}{169.87gAgNO_3} *\frac{1molAg_2S}{2molAgNO_3}*\frac{247.8gAg_2S}{1molAg_2S}\\\\m_{Ag_2S}^{theoretical}=3.11gAg_2S

Next, we compute the percent yield as:

Y=\frac{m_{Ag_2S}^{actual}}{m_{Ag_2S}^{theoretical}}*100\% =\frac{2.86g}{3.11g} *100\%\\\\Y=91.0%

Hence, answer is D. 91.0%.

Best regards.

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The answer to question 2
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Answer:


12


Explanation:


You will need a chemical equation with masses and molar masses, so let’s gather all the information in one place.


M_{r}:                           258.21       18.02


                 KAl(SO₄)₂·xH₂O ⟶ KAl(SO₄)₂ + xH₂O


Mass/g:             4.74                                       2.16


Step 1. Calculate the mass of the KAl(SO₄)₂.


Mass = 4.74 g – 2.16 g = 2.58 g.


Step 2. Calculate the moles of each product.


\text{Moles of KAl(SO}_{4})_{2} = \text{2.58 g} \times \frac{\text{1 mol} }{\text{258.21 g}} = 9.992 \times 10^{-3} \text{ mol}

\text{Moles of H}_{2}\text{O} = \text{2.16 g} \times \frac{\text{1 mol} }{\text{18.02 g}} = \text{ 0.1200 mol}

Step 3. Calculate the molar ratio of the two products.


\frac{\text{Moles of KAl(SO}_{4})_{2}}{\text{Moles of H}_{2}\text{O}} = \frac{ 9.992 \times 10^{-3} \text{ mol}}{\text{ 0.1200 mol} } = \frac{1 }{12.01} \approx \frac{ 1}{ 12}

1 mol of KAl(SO₄)₂ combines with 12 mol H₂O, so x = 12.



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