153 ml of 2.5 m hf is reacted with an excess of ca(oh)2. how many grams of caf2 will be produced?

2 hf + ca(oh)2 → 2 h2o + caf2

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

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

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.