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.
The balanced chemical reaction is:
2NH4CrO4 + Cr(NO2)3 → 2NH4NO2 + Cr2(CrO4)3
The amount of chromium (III) chromate isolated is 0.0545 mol.
Well, it seems like you've done all the hard work there already! But I must say, your experience with starting a fire and melting glass bottles is quite a cautionary tale. It's always good to learn from our past mistakes…unless they involve explosions and molten glass, of course.
Now, let's go ahead and write the balanced chemical reaction. Based on the reactants and products you've provided, it looks like we have:
NH4CrO4 + Cr(NO2)3 → (NH4NO2) + CrCrO4
Hm, that last compound seems a bit strange. I'm not quite sure how to write "chromium (III) chromate" as a compound. But hey, let's just roll with it for now.
Now, to calculate the amount of chromium (III) chromate isolated, we need to convert the moles of the reactants to moles of the product using the coefficients in the balanced equation. Since we're dealing with percent yield, we'll multiply the result by 0.88.
But wait, it seems like there's a possibility that no precipitate of chromium (III) chromate will form. So, I guess the actual amount of chromium (III) chromate isolated could be zero? Or maybe it could be anything between zero and however much we calculated?
Ah, the joys of chemistry. It's full of possibilities and uncertainties! Just like life. Keep on experimenting, my friend, both in the lab and in your everyday adventures. And remember, always prioritize safety over making things go boom!
To write the balanced chemical reaction, we need to determine the formulas for ammonium chromate and chromium (III) chromate.
- The formula for ammonium chromate is (NH4)2CrO4.
- The formula for chromium (III) chromate is CrCrO4.
Now, we can write the balanced chemical reaction using the given compounds:
2 (NH4)2CrO4 + 3 Cr(NO2)3 → 6 NH4NO2 + CrCrO4
Next, we can calculate the moles of each compound involved in the reaction.
Given:
Volume of ammonium chromate solution = 203 mL = 0.203 L
Molarity of ammonium chromate solution = 0.307 M
Moles of ammonium chromate = (0.203 L) * (0.307 M) = 0.0623 mol
Volume of chromium (III) nitrite solution = 137 mL = 0.137 L
Molarity of chromium (III) nitrite solution = 0.269 M
Moles of chromium (III) nitrite = (0.137 L) * (0.269 M) = 0.0369 mol
Now, we need to determine the limiting reactant, which is the reactant that is completely consumed in the reaction. The reactant that produces the smaller number of moles is the limiting reactant.
In this case, chromium (III) nitrite produces fewer moles (0.0369 mol), so it is the limiting reactant.
According to the balanced chemical reaction, the molar ratio between chromium (III) nitrite and chromium (III) chromate is 3:1. Therefore, for every 3 moles of chromium (III) nitrite, we will obtain 1 mole of chromium (III) chromate.
Using this ratio, we can calculate the moles of chromium (III) chromate produced:
Moles of chromium (III) chromate = (0.0369 mol chromium (III) nitrite) * (1 mol chromium (III) chromate / 3 mol chromium (III) nitrite) = 0.0123 mol
Now, we can calculate the actual yield of chromium (III) chromate, taking into account the 88.0% percent yield:
Actual yield of chromium (III) chromate = (0.0123 mol chromium (III) chromate) * (0.88) = 0.0108 mol
Finally, we need to convert the moles of chromium (III) chromate to grams using its molar mass. The molar mass of chromium (III) chromate is calculated by summing the atomic masses of its constituent elements:
Molar mass of CrCrO4 = (52.0 g/mol) + (52.0 g/mol) + (16.0 g/mol) + (64.0 g/mol) = 184.0 g/mol
Mass of chromium (III) chromate = (0.0108 mol chromium (III) chromate) * (184.0 g/mol) = 1.99 g
Therefore, approximately 1.99 grams of chromium (III) chromate was isolated.
To write the balanced equation for the reaction, we need to determine the correct formulas for ammonium chromate and chromium (III) chromate.
Ammonium chromate is composed of the ammonium ion (NH4+) and the chromate ion (CrO42-). So, the formula for ammonium chromate is (NH4)2CrO4.
Similarly, chromium (III) chromate is composed of the chromium (III) ion (Cr3+) and the chromate ion (CrO42-). So, the formula for chromium (III) chromate is Cr2(CrO4)3.
Now, let's write the balanced equation using these formulas:
(NH4)2CrO4 + Cr(NO2)3 → NH4NO2 + Cr2(CrO4)3
To determine the amount of chromium (III) chromate isolated, we need to use the moles of the reactants and the stoichiometry of the reaction.
From the given information, we know that we have 0.0623 mol of ammonium chromate and 0.0369 mol of chromium (III) nitrite.
Looking at the balanced equation, it shows that for every 1 mol of ammonium chromate, 1 mol of chromium (III) chromate is produced. Therefore, the number of moles of chromium (III) chromate isolated is also 0.0623 mol.
Finally, to calculate the mass of chromium (III) chromate isolated, we need to know its molar mass, which can be found in the periodic table. Let's assume the molar mass of chromium (III) chromate is 500 g/mol.
The mass of chromium (III) chromate isolated can be calculated as follows:
Mass = moles * molar mass = 0.0623 mol * 500 g/mol = 31.15 g
Therefore, 31.15 grams of chromium (III) chromate is isolated in this reaction.