Answer:
15 g
Explanation:
Data given:
amount of HF = 153 mL 2.5 M HF
amount of Ca(OH)₂ = Excess
grams of CaF₂ = ?
Reaction Given:
2HF + Ca(OH)₂ ------→ 2H₂O + CaF₂
Solution:
First we have to find number of moles of HF in 153 mL of 2.5 M HF
For this we will use following formula
Molarity = moles of solute / liter of solution
Rearrange above equation
moles of solute = Molarity x liter of solution . . . . . (1)
Put values in above equation (1)
moles of solute = 2.5 x 1 L
moles of solute = 2.5
So,
we come to know that there are 2.5 moles of solute (HF) in 1 L of solution
Now how many moles of solute will be present in 153 ml of solution
Convert 153 mL to Liter
1000 mL = 1 L
153 mL = 153/1000 = 0.153 L
Apply Unity Formula
2.5 moles HF ≅ 1 L solution
X moles of HF ≅ 0.153 L solution
moles of HF = 2.5 moles x 0.153 mL solution / 1 L solution
moles of HF = 0.383 moles
- So, 153 mL contains 0.383 moles of HF
Now Look at the reaction:
2HF + Ca(OH)₂ ------→ 2H₂O + CaF₂
2 mol 1 mol
From the reaction we come to know that 2 moles of HF gives 1 mole of CaF₂ then how many moles of CaF₂ will be produced from o.383 moles of HF
Apply Unity Formula
2 moles HF ≅ 1 mole of CaF₂
0.383 moles of HF ≅ X moles of CaF₂
moles of CaF₂ = 0.383 moles x 1 mole / 2 mol
moles of CaF₂ = 0.192 moles
- So, 0.192 moles of CaF₂ will be produced by 0.383 moles of HF
Now we will find mass of 0.192 moles of CaF₂
Formula will be used
mass in grams = no. of moles x molar mass . . . . . . . (2)
molar mass of CaF₂ = 40 + 2(19)
molar mass of CaF₂ = 40 + 38 = 78 g/mol
Put values in eq. 2
mass in grams = 0.192 x 78 g/mol
mass in grams = 14.976 g
rounding the value
mass in grams = 15 g
So,153 mL of 2.5 M HF is reacted with an excess of Ca(OH)₂ will produce 15 g of CaF₂.
