Answer :
Answer:
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