3C3H8 + 20H2CrO4-> 9CO2 + 10Cr2O3 + 32H2O
Suppose you want to make 1.50 g of chromium (III) oxide, how many ml of 6.25M chromic acid will you need to use if the percent yield of the reaction is 68.6%?
You want 1.5 g Cr2O3 so the question you ask yourself is, "What number must I start with to get 1.5 g if the reaction is only 68.6% efficient?"
That's 0.686*X = 1.5 and X = 1.5/0.686 = about 2.3g or so but you should do this more accurately.
Then convert 2.3(actually the number you obtain) to mols.
mols Cr2O3 = 2.3g/molar mass Cr2O3.
Next, using the coefficients in the balanced equation, convert mols Cr2O3 to mols H2CrO4.
Finally, M H2CrO4 = mols H2CrO4/L H2CrO4. You have M and mols solve for L and convert to mL.
how do I convert moles of Cr2O3 to moles of H2CrO4? I don't understand what this means
Use the coefficients in the balanced equation.
?mols Cr2O3 x (20 mols H2CrO4/10 mols Cr2O3)
To determine the volume of 6.25 M chromic acid needed, it's important to follow a step-by-step approach:
Step 1: Obtain the molar mass of chromium (III) oxide.
The molar mass of chromium (III) oxide (Cr2O3) can be calculated by adding the atomic masses of its constituent atoms: Cr (chromium) and O (oxygen). The atomic mass of chromium is 52 g/mol, and the atomic mass of oxygen is 16 g/mol. Since Cr2O3 contains two chromium atoms and three oxygen atoms, the molar mass is:
Molar mass of Cr2O3 = (2 x atomic mass of Cr) + (3 x atomic mass of O)
= (2 x 52 g/mol) + (3 x 16 g/mol)
= 104 g/mol + 48 g/mol
= 152 g/mol
Step 2: Calculate the amount of chromium (III) oxide required in moles.
Given that you want to make 1.50 g of chromium (III) oxide, you can convert this mass to moles using the molar mass obtained in Step 1. The equation to calculate moles is:
moles = mass / molar mass
moles of Cr2O3 = 1.50 g / 152 g/mol
≈ 0.00987 mol
Step 3: Determine the amount of chromic acid required.
By examining the balanced chemical equation, we can see that the molar ratio between chromic acid (H2CrO4) and chromium (III) oxide (Cr2O3) is 20 moles to 10 moles, or simplified as 2 moles to 1 mole.
The equation contains a 68.6% yield, meaning that 68.6% of the theoretical amount of chromium (III) oxide will be obtained. We can calculate the theoretical amount using the moles calculated in Step 2 and then determine the amount of chromic acid required as follows:
theoretical moles of Cr2O3 = 0.00987 mol
actual moles of Cr2O3 = theoretical moles of Cr2O3 x 0.686
Using the molar ratio, we can determine the required moles of chromic acid:
moles of H2CrO4 = (2/1) x actual moles of Cr2O3
Step 4: Convert the moles of chromic acid to volume.
Now, we need to convert the moles of chromic acid to volume using its molarity. The equation is:
volume (in liters) = moles / molarity
To convert the volume to milliliters, multiply by 1000:
volume (in mL) = volume (in L) x 1000
Finally, calculate the volume of chromic acid needed:
volume (in L) = moles of H2CrO4 / molarity of H2CrO4
volume (in mL) = volume (in L) x 1000
Insert the given molarity of 6.25 M for chromic acid and calculate the volume required.