2. Na2SiO3(s) + HF(aq) !H2SiF6(aq) + NaF(aq) + H2O (l)

a. Balance the reaction.

b. How many moles of HF are needed to react with 0.300 mol of Na2SiO3?

c. How many grams of NaF form when 0.500 mol of HF reacts with excess Na2SiO3?

d. How many grams of Na2SiO3can react with 0.800 g of HF?

e. Using your answer to part d-If you started with 1.5 g of Na2SiO3how many grams would be left after the reaction is complete?

a) Na2SiO3(s) + 8HF(aq) → H2SiF6(aq) + 2NaF(aq) + 3H2O (l)

b) 2.40 moles HF

c) 5.25 grams NaF

d)0.609 grams Na2SiO3

e) 0.89 grams Na2SiO3

Explanation:

Step 1: Data given

Step 2: The balanced equation

Na2SiO3(s) + 8HF(aq) → H2SiF6(aq) + 2NaF(aq) + 3H2O (l)

b) How many moles of HF are needed to react with 0.300 mol of Na2SiO3?

For 1 mol Na2SiO3 we need 8 moles HF to produce 1 mol H2SiF6, 2 moles NaF and 3 moles H2O

For 0.300 moles Na2SiO3 we'll need 8*0.300 = 2.40 moles HF

c. How many grams of NaF form when 0.500 mol of HF reacts with excess Na2SiO3?

For 1 mol Na2SiO3 we need 8 moles HF to produce 1 mol H2SiF6, 2 moles NaF and 3 moles H2O

For 0.500 moles we'll have 0.500 / 4 = 0.125 moles NaF

Mass NaF = 0.125 moles * 41.99 g/mol

Mass NaF = 5.25 grams NaF

d. How many grams of Na2SiO3 can react with 0.800 g of HF?

Moles HF = 0.800 grams / 20.01 g/mol

Moles HF = 0.0399 moles

For 1 mol Na2SiO3 we need 8 moles HF to produce 1 mol H2SiF6, 2 moles NaF and 3 moles H2O

For 8 moles we need 1 moles Na2SiO3 to react

For 0.0399 moles we need 0.0399/8 = 0.00499 moles Na2SiO3

Mass Na2SiO3 = 0.00499 moles * 122.06 g/mol  

Mass Na2SiO3 = 0.609 grams Na2SiO3

Using your answer to part d-If you started with 1.5 g of Na2SiO3how many grams would be left after the reaction is complete?

Number of moles HF = 0.0399 moles

Number of moles Na2SiO3 = 1.5 grams / 122.06 g/mol = 0.0123 moles

For 1 mol Na2SiO3 we need 8 moles HF to produce 1 mol H2SiF6, 2 moles NaF and 3 moles H2O

For 8 moles we need 1 moles Na2SiO3 to react

For 0.0399 moles we need 0.0399/8 = 0.00499 moles Na2SiO3

There will remain 0.0123 - 0.00499 = 0.00731 moles  Na2SiO3

Mass Na2SiO3 remaining = 0.00731 * 122.06 g/mol = 0.89 grams Na2SiO3

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole