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.

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.