The balanced chemical reaction is:
2NH4CrO4 + Cr(NO2)3 β 2NH4NO2 + Cr2(CrO4)3
The amount of chromium (III) chromate isolated is 0.0545 mol.
This is the problem: In a 1-L beaker, 203 ml of 0.307 M ammonium chromate was mixed with 137 ml of 0.269 M chromium (III) nitrite to produce ammonium nitrite and chromium (III) chromate. Write the balanced chemical reaction occuring here. If the percent yield of the reaction was 88.0%, how much chromium (III) chromate was isolated?
Work:
ammonium chromate: NH4 (how do you write chromate?)
chromium (III) nitrite: Cr(NO2)3
ammonium nitrite: NH4NO2
chromium (III) chromate: Cr (how do you write chromate?)
Molarity=moles of solute/liters of solution
moles of solute=(Molarity)(liters of solution)
203ml -> 0.203 L
(0.203L)(0.307 M) =.0623 mol
137ml -> 0.137 L
(0.137L)(0.269 M) = 0.0369 mol
Chromate ion is CrO4^-2. Perhaps that is all you need. Remember to convert mols of what you start with to mols of the end product using the coefficients in the balanced equation. The final 88% yield is calculated by
g end product x 0.88 = ??
I have been unable to find any indication that chromium(III) chromate is insoluble in water; therefore, I question if a precipitate of chromium(III) chromate will form. Frankly, I don't believe a reaction will take place to produce chromium(III) chromate. There is a distinct possibility of a redox reaction taking place in which the chromate will be reduced and the NO2^- will be oxidized. (That happened to me about 50 years ago and I started a fire hot enough to melt the glass bottle which I was holding in my hand and the molten glass was not kind to my hands). The experience is indelibly etched in my mind. But all this aside, I think we work the problem as if a ppt of chromium(III) chromate is formed.
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