How many grams of commercial C2H4O2 (97% pure by mass) must be allowed to react with an excess of PCl3 to produce 75.0g of C2H3OCL if the reaction has a 78.2% yield?
C2H4O2 = A with molar mass approx 60.
C2H3OCl = B with molar mass approx 78,
Note that all of these numbers are estimates so you should recalculate throughout for more accurate values.
A + PCl3 = B
Since the yield is 78.2% and not 100%, you will need to prepare 75.0/0.782= approx 95 g in order to obtain 75 g.
mols B = 95/78 = about 1.2
1 mol B is obtained from 1 mol A; therefore, you must start with 1.2 mols A.
Convert mols A to grams A by 1.2 mols A x 60 = approx 73 g but that's true only for 100% A and it is only 97% pure so you must start with 73/0.97 = approx ? g.
To determine the amount of commercial C2H4O2 needed to produce 75.0g of C2H3OCL, we need to follow these steps:
Step 1: Calculate the molar mass of C2H3OCL.
The molar mass of C2H3OCL is equal to the sum of the molar masses of its individual atoms.
C = 12.01 g/mol
H = 1.01 g/mol
O = 16.00 g/mol
Cl = 35.45 g/mol
Total molar mass of C2H3OCL = 2(12.01) + 3(1.01) + 16.00 + 35.45 = 78.14 g/mol
Step 2: Calculate the moles of C2H3OCL.
We can use the given mass of C2H3OCL (75.0g) and its molar mass to calculate the number of moles.
Moles of C2H3OCL = mass / molar mass = 75.0g / 78.14 g/mol ≈ 0.959 mol
Step 3: Calculate the moles of C2H4O2 required.
Since the reaction is not 100% yield, we need to account for the percent yield. The percent yield is given as 78.2%.
So, the actual moles of C2H3OCL produced is 78.2% of the calculated moles obtained in Step 2.
Actual moles of C2H3OCL produced = 0.782 × 0.959 mol ≈ 0.750 mol
Step 4: Calculate the moles of PCl3 consumed.
The balanced chemical equation between C2H4O2 and PCl3 is:
C2H4O2 + PCl3 → C2H3OCL + HCl
From the equation, we can see that the moles of C2H4O2 and PCl3 are in a 1:1 ratio.
Therefore, moles of PCl3 consumed = moles of C2H3OCL produced = 0.750 mol
Step 5: Calculate the grams of commercial C2H4O2.
Since the commercial C2H4O2 is 97% pure by mass, it means that 97% of the mass is due to C2H4O2, and the remaining 3% is impurities.
To find the mass of pure C2H4O2 required, we can set up the following equation:
0.750 mol C2H4O2 × (molar mass of C2H4O2 / 1 mol) × (mass percent of pure C2H4O2 / 100%)
= 0.750 mol × (60.05g/mol) × (97% / 100%)
≈ 43.60g
Therefore, approximately 43.60 grams of commercial C2H4O2 must be allowed to react with an excess of PCl3 to produce 75.0 grams of C2H3OCL if the reaction has a 78.2% yield.
To solve this problem, we need to first determine the balanced chemical equation for the reaction between C2H4O2 and PCl3. From the given information, we know that the reaction produces C2H3OCl.
1. Write the balanced chemical equation:
C2H4O2 + PCl3 -> C2H3OCl + HCl
Next, calculate the molecular weight of C2H3OCl to determine the number of moles produced.
2. Calculate the molecular weight of C2H3OCl:
C (12.01 g/mol) + H (1.007 g/mol) + Cl (35.45 g/mol) = 48.47 g/mol
Since 1 mole of C2H3OCl has a molecular weight of 48.47 g/mol, we can calculate the number of moles of C2H3OCl produced from the given mass.
3. Calculate the number of moles of C2H3OCl produced:
Mass of C2H3OCl / Molecular weight = 75.0 g / 48.47 g/mol ≈ 1.55 mol
Now, let's calculate how many moles of C2H4O2 are required to produce this amount of C2H3OCl while considering the reaction's yield.
4. Calculate the number of moles of C2H4O2 required:
Number of moles of C2H3OCl / Yield = 1.55 mol / 0.782 ≈ 1.98 mol
Since the reaction between C2H4O2 and PCl3 occurs in a 1:1 stoichiometric ratio, we can conclude that 1.98 moles of C2H4O2 are required.
Finally, we can calculate the mass of commercial C2H4O2 (97% pure) needed, considering its percentage purity by mass.
5. Determine the mass of commercial C2H4O2 needed:
Mass of commercial C2H4O2 = (Number of moles of C2H4O2) x (Molecular weight of C2H4O2 / Purity Percentage)
Let's calculate the mass:
Mass of commercial C2H4O2 = 1.98 mol x (60.05 g/mol / 0.97) ≈ 122.63 g
Therefore, approximately 122.63 grams of commercial C2H4O2 (97% pure by mass) must be allowed to react with an excess of PCl3 to produce 75.0 grams of C2H3OCl if the reaction has a 78.2% yield.