Chem question of wrath (1 Viewer)

[Real Madrid]

A silver coin was analysed and found to contain only silver and copper. 1.580 g of this coin was dissolved in concentrated nitric acid and the resultant solution diluted. Reaction of the solution with excess hyderochloric acid yielded 1.050 g of silver chloride. Calculate the percentage of silver in the sample.


Having trouble answering

 

[bored of sc]

Not Enough Information?

Just a question: Is the copper and silver alloy a compound or mixture? The whole alloy thing is confusing.

 

[Real Madrid]

Its a coin so so assume it is an alloy. I've caluculated the reactions but I cannot find the percentages.

2Ag+2HNO3--->2Ag(NO3)+H2

2Ag(NO3)+2HCl--->2AgCl+2HNO3

The amount of SILVER(I) in the coin has been calculated to be 1.580676666 grams by dividing the Number Of Moles By X in each compound, using the formula X = 2867/21 = 136.5238095.

I don't know how to figure percentages.

 

[bored of sc]

Don't you need to find the mass of the copper in the coin to work out the percentage?

 

[Finx]

Sheet 12/14 or something right?

Our teacher told us not to do this one because it wasn't worded properly.

 

[bored of sc]

What topic is this in by the way? Metals? What dot-point of the syllabus?

 

[minijumbuk]

I answered this question a while ago in another thread.

2Ag(s) + 2HNO3(aq) --> 2AgNO3(aq) + H2(g)
AgNO3(aq) + HCl(aq) --> AgCl(s) + HNO3(aq)

Cu is a weaker reducing agent than H+, so will not react


Moles of AgCl = 1.05 g / (107.9 +35.45) g/mol
= X mol (press this on the calculator yourself xD)

Therefore X mol AgNO3 reacted (molar ratio 1:1)
And hence, X mol Ag reacted in first equation to form X mol AgNO3 (molar ratio of Ag : AgNO3 = 2:2 = 1:1)

Hence, Mass of Ag = X mol x 107.9

Therefore mass percentage of Silver in the coin mixture = (107.9 x X) / 1.58 x 100%

 

[lolokay]

..can't you just find % of Ag in AgCl =~75%. 0.75*1.05 =~0.79. 0.79/1.58 = 0.5, so 50%?

edit: nevermind, that was basically what minijumbuk did anyway