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?
I answered your earlier post below.
To find the mass of the solid product that was isolated, we can start by calculating the number of moles of each reactant using the given volumes and concentrations. Then, we can determine the limiting reactant and use stoichiometry to find the number of moles of the product formed. Finally, we can convert the moles of product to mass using the molar mass of the product.
Step 1: Calculate the number of moles of each reactant.
- Ammonium chromate (NH4)2CrO4:
To find the number of moles, we will use the formula:
moles = volume (in liters) * concentration (in moles per liter)
moles of (NH4)2CrO4 = 203 mL * (1 L / 1000 mL) * 0.307 mol/L = 0.062361 mol
- Chromium(III) nitrite Cr(NO2)3:
moles of Cr(NO2)3 = 137 mL * (1 L / 1000 mL) * 0.269 mol/L = 0.036853 mol
Step 2: Determine the limiting reactant.
The limiting reactant is the reactant that is completely consumed in a reaction, limiting the amount of product that can be formed. To find the limiting reactant, we compare the number of moles of each reactant and determine which is smaller.
From step 1, we have:
moles of (NH4)2CrO4 = 0.062361 mol
moles of Cr(NO2)3 = 0.036853 mol
Since the coefficients in the balanced equation are 3:2 for (NH4)2CrO4 and Cr(NO2)3, respectively, we can see that Cr(NO2)3 is the limiting reactant.
Step 3: Use stoichiometry to find the number of moles of the product.
From the balanced equation:
3 (NH4)2CrO4 + 2 Cr(NO2)3 -> Cr2(CrO4)3 + 6 NH4NO2
Since the stoichiometric coefficient for Cr2(CrO4)3 is 1, the number of moles of Cr2(CrO4)3 formed will be equal to the number of moles of Cr(NO2)3.
Therefore, moles of Cr2(CrO4)3 = 0.036853 mol
Step 4: Convert moles of product to mass using the molar mass of the product.
The molar mass of Cr2(CrO4)3 is the sum of the atomic masses of its elements:
Molar mass of Cr2(CrO4)3 = (2 * atomic mass of Cr) + (3 * atomic mass of O)
Using the periodic table, the atomic mass of Cr is approximately 52 g/mol, and the atomic mass of O is approximately 16 g/mol.
Molar mass of Cr2(CrO4)3 = (2 * 52 g/mol) + (3 * 16 g/mol) = 204 g/mol
Finally, the mass of the product can be found using the formula:
mass = moles * molar mass
mass of Cr2(CrO4)3 = 0.036853 mol * 204 g/mol = 7.512 g
Therefore, the mass of the solid product isolated is approximately 7.512 grams.