Chem Calcualtions 3/4

[justsoundslikeaworn-outcliche]

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?

[kaiipoo_]

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 (CrO4^2-) 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(CrO₄^2-)

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.

[justsoundslikeaworn-outcliche]

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 (CrO4^2-) 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(CrO₄^2-)

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

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

[Ancora_Imparo]

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)

[justsoundslikeaworn-outcliche]

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.