ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: justsoundslikeaworn-outcliche on March 12, 2013, 08:25:39 pm
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So I posted this before as a photo but I wasn't getting many replies since lets be honest no one opens attachments. Anyway, I have 2 questions:
When solution of silver nitrate and potassium dichromate react they produce a precipitate of silver dichromate. Balnced Equation: 2AgNO3 + K2CrO4 -> Ag2CrO4 + 2KNO3
If 0.788g of precipitate forms in the reaction find:
i)mass of dichromate that reacted
ii)mass of silver nitrate that reacted
iii)percentage by mass of silver in the precipitate?
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This is a stoichiometry based question and thus you have to work with the ratios.
2AgNO3 + K2CrO4 -> Ag2CrO4 + 2KNO3
Q) If 0.788g of precipitate forms in the reaction find:
i)mass of dichromate that reacted
ii)mass of silver nitrate that reacted
iii)percentage by mass of silver in the precipitate?
A) i) Find the amount of precipitate in mol. that is, n = m/M.
According to the equation, the amount of chromate (CrO42-) is in a 1:1 mol ratio.
so the amount of chromate in mol is equal to the amount of precipitate.
To find the mass, rearrange the mol equation so m= nM, and sub.
ii) Same as i), except silver nitrate reacts in a 2:1 ratio, so your number of mol of Silver Nitrate will be 2x n(CrO42-)
iii) Percentage by Mass of Silver in Ag2CrO4 (M= 331.8 g mol-1)
= molar mass of silver/ molar mass of silver chromate.
which is effectively 215.8/331.8 = 65.0392%.
you have a precipitate of 0.788 grams, so the percentage of mass by silver will be 65.0392% of that.
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This is a stoichiometry based question and thus you have to work with the ratios.
2AgNO3 + K2CrO4 -> Ag2CrO4 + 2KNO3
Q) If 0.788g of precipitate forms in the reaction find:
i)mass of dichromate that reacted
ii)mass of silver nitrate that reacted
iii)percentage by mass of silver in the precipitate?
A) i) Find the amount of precipitate in mol. that is, n = m/M.
According to the equation, the amount of chromate (CrO42-) is in a 1:1 mol ratio.
so the amount of chromate in mol is equal to the amount of precipitate.
To find the mass, rearrange the mol equation so m= nM, and sub.
ii) Same as i), except silver nitrate reacts in a 2:1 ratio, so your number of mol of Silver Nitrate will be 2x n(CrO42-)
iii) Percentage by Mass of Silver in Ag2CrO4 (M= 331.8 g mol-1)
= molar mass of silver/ molar mass of silver chromate.
which is effectively 215.8/331.8 = 65.0392%.
you have a precipitate of 0.788 grams, so the percentage of mass by silver will be 65.0392% of that.
THANKYOU SO MUCH!! you really cleared up a lot for me :D
One more question sorry hah :
Mineral water with an unknown amount of Potassium carbonate, is taken and 20ml aliquots of this solution are titred against 0.15M HCL, an average titre=12.30ml is found
how do I find the molarity of potassium carbonate?
AND what would be the mass of potassium carbonate present in 1L of mineral water?
Thanks heaps :D
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K2CO3 + 2HCl -> H2CO3 + 2KCl
For molarity:
n(HCl) = c * V = 0.15 * 0.01230
n(K2CO3 in aliquot) = 1/2 * n(HCl) [from equation above]
c(K2CO3) = n/v = n(K2CO3 in aliquot)/0.020
For mass in 1L:
n(K2CO3 in 1L) = c(K2CO3)
m(K2CO3 in 1L) = n * M = n(K2CO3 in 1L) * M(K2CO3)
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K2CO3 + 2HCl -> H2CO3 + 2KCl
For molarity:
n(HCl) = c * V = 0.15 * 0.01230
n(K2CO3 in aliquot) = 1/2 * n(HCl) [from equation above]
c(K2CO3) = n/v = n(K2CO3 in aliquot)/0.020
For mass in 1L:
n(K2CO3 in 1L) = c(K2CO3)
m(K2CO3 in 1L) = n * M = n(K2CO3 in 1L) * M(K2CO3)
Thanks man your a legend.