CaSiO3(s) +6 HF (g)-> CaF2(aq) + SiF4 (g) + 3 H2O(l)
If a 65.6 g sample of CaSiO3 is reacted in an inert 83 L container at 25.5°C with HF at 1.00 atm pressure find the following.
a. the mass of water, and CaF2 produced in the reaction and the mass of HF that remains.
b. The partial pressure of SiF4 in the container after the temperature has returned to 25.5°C
Where are you stuck on this problem?
mols CaSiO3 = grams/molar mass = ?
Use PV = nRT to find mols HF = ?
Find limiting reagent.
I found 3.39 mols of HF and 0.56 mols of CaSiO3 but I'm confused what is the limiting reactant. It would be CaSio3 because it has less moles than HF?
To solve this problem, we'll follow these steps:
Step 1: Convert grams of CaSiO3 to moles.
Step 2: Use the balanced chemical equation to determine the mole ratios between CaSiO3, CaF2, and H2O.
Step 3: Convert moles of CaF2 and H2O to grams.
Step 4: Determine the moles of HF consumed in the reaction.
Step 5: Calculate the moles of HF remaining.
Step 6: Use the ideal gas law to calculate the partial pressure of SiF4.
Let's go through each step in detail:
Step 1: Convert grams of CaSiO3 to moles.
We'll use the molar mass of CaSiO3 to convert the given mass to moles.
Molar mass of CaSiO3: Ca (40.08 g/mol) + Si (28.09 g/mol) + 3O (16.00 g/mol) = 116.27 g/mol
Moles of CaSiO3 = Mass of CaSiO3 / Molar mass of CaSiO3
Moles of CaSiO3 = 65.6 g / 116.27 g/mol
Moles of CaSiO3 ≈ 0.564 mol
Step 2: Use the balanced chemical equation to determine the mole ratios.
From the balanced equation:
1 mol CaSiO3 produces 1 mol CaF2 and 3 mol H2O.
Step 3: Convert moles of CaF2 and H2O to grams.
Molar mass of CaF2: Ca (40.08 g/mol) + 2F (19.00 g/mol) = 78.08 g/mol
Mass of CaF2 = Moles of CaF2 x Molar mass of CaF2
Mass of CaF2 = 0.564 mol x 78.08 g/mol
Mass of CaF2 ≈ 44.03 g
Molar mass of H2O: 2H (2.02 g/mol) + O (16.00 g/mol) = 18.02 g/mol
Mass of H2O = Moles of H2O x Molar mass of H2O
Mass of H2O = 3 mol x 18.02 g/mol
Mass of H2O ≈ 54.06 g
Step 4: Determine the moles of HF consumed in the reaction.
Using the mole ratio from the balanced equation, we find that 1 mole of CaSiO3 reacts with 6 moles of HF.
Moles of HF consumed = 0.564 mol x 6 = 3.384 mol
Step 5: Calculate the moles of HF remaining.
To find the moles of HF remaining, we need to know the initial moles of HF present in the 83 L container at 1.00 atm and 25.5°C.
Assuming an ideal gas, we can use the ideal gas law:
PV = nRT
R is the ideal gas constant (0.0821 L·atm/mol·K).
T should be expressed in Kelvin, so 25.5°C + 273.15 = 298.65 K.
PV = nRT
(1.00 atm) x (83.0 L) = n x (0.0821 L·atm/mol·K) x (298.65 K)
n = (1.00 atm x 83.0 L) / (0.0821 L·atm/mol·K x 298.65 K)
n ≈ 3.370 mol (initial moles of HF)
Moles of HF remaining = Initial moles of HF - Moles of HF consumed
Moles of HF remaining = 3.370 mol - 3.384 mol
Moles of HF remaining ≈ -0.014 mol (Negative value means HF is consumed completely)
Step 6: Use the ideal gas law to calculate the partial pressure of SiF4.
We'll use the ideal gas law again to calculate the partial pressure of SiF4 since there is no specific information given about it.
PV = nRT
P = (n/V) x RT
From the balanced equation, 1 mole of CaSiO3 reacts to form 1 mole of SiF4.
Moles of SiF4 = 0.564 mol
P( SiF4) = (0.564 mol / 83.0 L) x (0.0821 L·atm/mol·K) x (298.65 K)
P( SiF4) ≈ 0.301 atm
So, the partial pressure of SiF4 in the container after the temperature has returned to 25.5°C is approximately 0.301 atm.