The reaction of 9.5 grams of fluorine with
excess chlorine produced 5.9 grams of ClF3.
What percent yield of ClF3 was obtained?
Answer in units of %
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To calculate the percent yield, we need to compare the actual yield (5.9 grams of ClF3) to the theoretical yield.
Step 1: Calculate the molar mass of ClF3:
- The molar mass of Cl is 35.45 grams/mol.
- The molar mass of F is 18.99 grams/mol.
- Since ClF3 has one Cl atom and three F atoms, the molar mass of ClF3 is:
(1 * 35.45 grams/mol) + (3 * 18.99 grams/mol) = 96.42 grams/mol.
Step 2: Convert the mass of fluorine (9.5 grams) to moles:
- To do this, divide the given mass by the molar mass of F:
9.5 grams ÷ 18.99 grams/mol = 0.50 moles of F.
Step 3: Apply stoichiometry to determine the theoretical yield:
- The balanced chemical equation for the reaction between fluorine (F2) and chlorine (Cl2) to form ClF3 is:
F2 + 3Cl2 -> 2ClF3.
- From the equation, we can see that 1 mole of F2 reacts with 3 moles of Cl2 to produce 2 moles of ClF3.
- Therefore, the moles of ClF3 produced can be calculated as follows:
(0.50 moles of F) * (2 moles of ClF3 / 1 mole of F2) = 1.00 mole of ClF3.
Step 4: Calculate the theoretical yield in grams:
- To convert moles of ClF3 to grams, multiply the moles by the molar mass of ClF3:
1.00 mole of ClF3 * 96.42 grams/mol = 96.42 grams of ClF3.
Step 5: Calculate the percent yield:
- Divide the actual yield (5.9 grams) by the theoretical yield (96.42 grams) and multiply by 100 to get the percent yield:
(5.9 grams / 96.42 grams) * 100 = 6.11%.
Therefore, the percent yield of ClF3 obtained is approximately 6.11%.