How many grams of silver chromate will precipitate when 150 mL of 0.500 M silver nitrate are added to 100 mL of 0.400 M potassium chromate?

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How many grams of silver chromate will precipitate when 150 mL of 0.500 M silver nitrate are added to 100 mL of 0.400 M potassium chromate?

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Answer 1 · 2017-04-23T05:43:15

WE need (i) a stoichiometric equation........and should get over $12 \cdot g$ $12*g$ of a brick-red precipitate.

Explanation:

$2 A g^{+} + C r O_{4}^{2 -} \to A g_{2} C r O_{4} (s) ⏐ ⏐ ↓$ $2Ag^(+) + CrO_4^(2-) rarr Ag_2CrO_4(s)darr$

Silver chromate is EXCEPTIONALLY insoluble, and will deposit with alacrity as a brick-red precipitate.

And we need equivalent quantities of silver ion and chromate ion.

$Moles of silver ion = 150 \times 10^{- 3} \cdot L \times 0.500 \cdot m o l \cdot L^{- 1} = 0.075 \cdot m o l$ $"Moles of silver ion"=150xx10^-3*Lxx0.500*mol*L^-1=0.075*mol$

$Moles of chromate ion = 100 \times 10^{- 3} \cdot L \times 0.400 \cdot m o l \cdot L^{- 1} = 0.040 \cdot m o l$ $"Moles of chromate ion"=100xx10^-3*Lxx0.400*mol*L^-1=0.040*mol$

Clearly, there is EXCESS chromate ion, MORE than 1/2 equiv. And thus all the silver ion will precipitate as silver chromate.

Given the stoichiometry, $\frac{0.075}{2} \cdot m o l$ $0.075/2*mol$ $A g_{2} C r O_{4}$ $Ag_2CrO_4$ (approx. $12 \cdot g)$ $12*g)$ should precipitate......

i.e. $0.0375 \cdot m o l \times 331.73 \cdot g \cdot m o l^{- 1} = ? ? \cdot g$ $0.0375*molxx331.73*g*mol^-1=??*g$

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How many grams of silver chromate will precipitate when 150 mL of 0.500 M silver nitrate are added to 100 mL of 0.400 M potassium chromate?

1 Answer

Explanation:

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