Hydrogen fluoride is produced from the reaction of calicum fluoride with sulfuric acid, which also forms calicum sulfate. 6.00 kg of calicum fluoride are reacted with an excess of sulfuric acid. How much hyrdogen fluoride is formed if the percent yield is 93.0%?
Here is an example. Just follow the steps. http://www.jiskha.com/science/chemistry/stoichiometry.html
Multiply the theoretical yield you find by 0.93.
is the theoretical yield 6.00?
No. Where did you get the 6.00? From the problem of 6.00 kg? You must go through the procedure outlined in the link to find the theoretical yield.
To calculate the amount of hydrogen fluoride (HF) formed, we need to use the given information and follow these steps:
Step 1: Write the balanced chemical equation for the reaction:
CaF2 + H2SO4 → CaSO4 + 2HF
According to the equation, 1 mole of calcium fluoride (CaF2) reacts to form 2 moles of hydrogen fluoride (HF).
Step 2: Determine the molar mass of CaF2 and HF:
Molar mass of CaF2: 40.08 g/mol (calcium) + 2 * 19.00 g/mol (fluorine) = 78.08 g/mol
Molar mass of HF: 1.01 g/mol (hydrogen) + 19.00 g/mol (fluorine) = 20.01 g/mol
Step 3: Calculate the number of moles of CaF2:
Using the given mass and molar mass of CaF2:
6.00 kg × (1000 g / 1 kg) ÷ (78.08 g/mol) = 76.88 mol
Step 4: Determine the maximum amount of HF that would be produced:
Since 1 mole of CaF2 forms 2 moles of HF, the theoretical yield of HF is double the moles of CaF2:
76.88 mol × 2 = 153.76 mol
Step 5: Calculate the actual yield of HF:
Using the percent yield of 93.0%, we multiply the maximum theoretical yield by the decimal form of the percentage:
93.0% = 93.0 ÷ 100 = 0.93
Actual yield = 0.93 × 153.76 mol = 142.99 mol
Step 6: Convert the moles of HF into grams:
Using the molar mass of HF:
142.99 mol × 20.01 g/mol = 2,861.99 g ≈ 2.86 kg
Therefore, approximately 2.86 kg of hydrogen fluoride is formed if the percent yield is 93.0%.