Question #7e402

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Question #7e402

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Answer 1 · 2015-06-18T21:39:54

19.49g

Explanation:

$2 A g F_{(a q)} + N a_{2} S O_{4 (a q)} \to A g_{2} S O_{4 (s)} + 2 N a F_{(a q)}$ $2AgF_((aq))+Na_2SO_(4(aq))rarrAg_2SO_(4(s))+2NaF_((aq))$

$c = \frac{n}{v}$ $c=n/v$

$n = c v$ $n=cv$

So the number of moles of $A g F =$ $AgF=$

$\frac{2.5 \times 50.0}{1000} = 0.125$ $(2.5xx50.0)/(1000)=0.125$

From the equation we can see that 2 moles of $A g F$ $AgF$ produces 1 mole of $A g_{2} S O_{4}$ $Ag_2SO_4$

So the no. moles $A g_{2} S O_{4} = \frac{0.125}{2} = 0.0625$ $Ag_2SO_4=0.125/2=0.0625$

The $M r$ $Mr$ of $A g_{2} S O_{4} = 311.8$ $Ag_2SO_4=311.8$

So the mass of $A g_{2} S O_{4} = 311.8 \times 0.0625 = 19.49 g$ $Ag_2SO_4=311.8xx0.0625=19.49"g"$

Answer 2 · 2015-06-18T21:53:07

The mass of the precipitate will be equal to 19.5 g.

Explanation:

So, you're two solutions that contain soluble compounds - silver fluoride, $A g F$ $AgF$ , and sodium sulfate, $N a_{2} S O_{4}$ $Na_2SO_4$ .

The double replacement reaction that takes place between these two compounds will produce silver sulfate, $A g_{2} S O_{4}$ $Ag_2SO_4$ , an insoluble compound, and sodium fluoride, $N a F$ $NaF$ , according to the balanced chemical equation

$2 A g F_{(a q)} + N a_{2} S O_{4 (a q)} \to A g_{2} S O_{4 (s)} ⏐ ↓ + 2 N a F_{(a q)}$ $color(red)(2)AgF_((aq)) + Na_2SO_(4(aq)) -> Ag_2SO_(4(s)) darr + 2NaF_((aq))$

Since sodium sulfate is in excess, all the moles of silver fluoride will react. The $2 : 1$ $color(red)(2):1$ mole ratio that exists between silver fluoride and silver sulfate tells you that, regardless of how many moles of the former react, the reaction will produce half as many moles of the latter.

Use the molarity and volume of the silver fluoride solution to determine how many moles take part in the reaction

$C = \frac{n}{V} \Rightarrow n = C \cdot V$ $C = n/V => n = C * V$

$n_{A g F} = 2.50 M \cdot 50.0 \cdot 10^{- 3} L = 0.125 moles$ $n_(AgF) = "2.50 M" * 50.0 * 10^(-3)"L" = "0.125 moles"$ $A g F$ $AgF$

This means that the reaction will produce

$0.125cancel("moles"AgF) * ("1 mole "Ag_2SO_4)/(color(red)(2)cancel("moles"AgF)) = "0.0625 moles"$ $Ag_2SO_4$

To get the mass of the precipitate, use its molar mass

$0.0625cancel("moles") * "311.8 g"/(1cancel("mole")) = color(green)("19.5 g")$

The net ionic equation for this reaction looks like this

$2Ag_((aq))^(+) + SO_(4(aq))^(2-) -> Ag_2SO_(4(s)) darr$

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Question #7e402

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