In a 1L beaker, 203mL of 0.307 M ammonium chromate was mixed with 137mL of 0.269 M chromium(iii) nitrite.
3(NH4)2CrO4 + 2Cr(NO2)3 -> Cr2(CrO4)3 + 6NH4NO2
If the percent yield of the reaction was 88.0 %, what mass of solid product was isolated?
Anytime amounts are given for BOTH reactants you know it is a limiting reagent (LR) problem and the first thing you must do is determine the LR.
mols (NH4)2CrO4 = M x L = ?
mols Cr(NO3)3 = M x L = ?
Using the coefficients in the balanced equation, convert mols EACH one into mols of the product. It is likely that the two values will not agree; the correct value in LR problems is ALWAYS the smaller value and the reagent producing that number is the LR.
Use the smaller value of mols for the product and convert that to grams. g = mols x molar mass. This the theoretical yield (that is what you would get at 100%).
Then since the % yield is only 88.0%, multily the grams by 0.88. That is the grams of product you will actually receive.
To calculate the mass of solid product isolated, you first need to determine the limiting reagent. The limiting reagent is the reactant that is completely consumed in a chemical reaction, limiting the amount of product that can be formed.
1. Calculate the number of moles of ammonium chromate (NH4)2CrO4:
Number of moles = Volume (in L) x Concentration (in mol/L)
Moles of (NH4)2CrO4 = 203 mL x 0.307 mol/L = 0.062261 mol
2. Calculate the number of moles of chromium (III) nitrite Cr(NO2)3:
Moles of Cr(NO2)3 = 137 mL x 0.269 mol/L = 0.036953 mol
3. Determine the stoichiometry of the reaction:
3(NH4)2CrO4 + 2Cr(NO2)3 -> Cr2(CrO4)3 + 6NH4NO2
From the balanced equation, you can see that the stoichiometric ratio between (NH4)2CrO4 and Cr(NO2)3 is 3:2. This means that 3 moles of (NH4)2CrO4 react with 2 moles of Cr(NO2)3.
4. Determine the limiting reagent:
To determine the limiting reagent, compare the moles of each reactant to the stoichiometric ratio.
Moles of (NH4)2CrO4 / Stoichiometric coefficient of (NH4)2CrO4 = 0.062261 mol / 3 = 0.020754 mol
Moles of Cr(NO2)3 / Stoichiometric coefficient of Cr(NO2)3 = 0.036953 mol / 2 = 0.018477 mol
The limiting reagent is Cr(NO2)3 because it produces fewer moles of product.
5. Calculate the theoretical yield of Cr2(CrO4)3:
Theoretical yield of Cr2(CrO4)3 = Moles of limiting reagent x Molar mass of Cr2(CrO4)3
Molar mass of Cr2(CrO4)3 = (52.00 g/mol + 16.00 g/mol)*3 = 216 g/mol
Theoretical yield of Cr2(CrO4)3 = 0.018477 mol x 216 g/mol = 3.995 SE
6. Calculate the mass of solid product isolated:
Mass of solid product isolated = Theoretical yield x Percent yield
Mass of solid product isolated = 3.995 g x 0.880 = 3.5166 g
Therefore, the mass of solid product isolated is approximately 3.5166 grams.
To find the mass of the solid product isolated, we need to first calculate the moles of each reactant, then determine the limiting reactant, and finally use the stoichiometry of the balanced equation to find the moles and mass of the solid product.
Let's go step by step:
1. Calculate the moles of ammonium chromate (NH4)2CrO4:
Moles of (NH4)2CrO4 = volume (in liters) x molarity
The volume is 203 mL, which is 0.203 L.
Moles of (NH4)2CrO4 = 0.203 L x 0.307 mol/L
2. Calculate the moles of chromium(III) nitrite Cr(NO2)3:
Moles of Cr(NO2)3 = volume (in liters) x molarity
The volume is 137 mL, which is 0.137 L.
Moles of Cr(NO2)3 = 0.137 L x 0.269 mol/L
3. Determine the limiting reactant:
The limiting reactant is the one that is completely consumed and determines the maximum amount of product that can be formed. To find the limiting reactant, we need to compare the moles of each reactant obtained in step 1 and 2.
The coefficients in the balanced equation show that 3 moles of (NH4)2CrO4 react with 2 moles of Cr(NO2)3. So, we need to compare the moles of (NH4)2CrO4 and Cr(NO2)3 using these ratios.
moles of (NH4)2CrO4 = 0.203 L x 0.307 mol/L = 0.062361 mol
moles of Cr(NO2)3 = 0.137 L x 0.269 mol/L = 0.036953 mo
From the stoichiometry, it can be observed that 3 moles of (NH4)2CrO4 are required for every 2 moles of Cr(NO2)3. So, let's compare the moles in that ratio:
(moles of (NH4)2CrO4) / 3 = (moles of Cr(NO2)3) / 2
(0.062361 mol) / 3 = (0.036953 mol) / 2
The ratio is about 1.04. Since we can't have a partial moles, we can conclude that Cr(NO2)3 is the limiting reactant.
4. Calculate the moles of the solid product Cr2(CrO4)3:
Using the stoichiometry of the balanced equation, we can see that for every 2 moles of Cr(NO2)3, we obtain 1 mole of Cr2(CrO4)3.
Moles of Cr2(CrO4)3 = (moles of Cr(NO2)3) x (1 mol Cr2(CrO4)3 / 2 mol Cr(NO2)3)
Moles of Cr2(CrO4)3 = (0.036953 mol) x (1 mol / 2 mol)
5. Calculate the mass of the solid product Cr2(CrO4)3:
Mass of Cr2(CrO4)3 = moles of Cr2(CrO4)3 x molar mass of Cr2(CrO4)3
The molar mass of Cr2(CrO4)3 can be calculated by adding up the atomic masses of all the atoms in one formula unit.
Finally, to find the mass of the solid product isolated, multiply the moles obtained in step 4 by the molar mass of Cr2(CrO4)3.
Please note that the balanced equation and molar masses used in this explanation may not be accurate as they are not provided in the original question.