Question

Please note: There are 3 questions listed, but each question has 3 sub-questions. Thank you :)

7. How many grams of sodium cabonate will react with 25.5mL of 0.300 M hydrochloric acid?

Na₂CO₃ + HCl -> NaCl + H₂O + CO₂ (unbalanced)

a. determine the number of mole (mmoles) hydrochloric acid used.

b. Determine the number of moles (mmoles) of sodium carbonate reacted.

c. Determine the mass (grams) of sodium cobonate reacted.

8. How many grams of aluminum hydroxide will react with 75.5 mL of 0.250 M sulfuric acid?

Al(OH)₃ + H₂SO₄ -> Al₂(SO₄)₃ + H₂O (unbalanced)

a. Determine the number of mole (mmoles) sulfuric acid used.

b. Determine the number of moles (mmoles) Aluminum hydroxide reacted.

c. Determine the mass (grams) of aluminum hydroxide reacted..

9. Consider the following solutioins. Put in order according to the boiling points from highest point to lowest. (Hint: Colligative properties. k_b = 0.52C /m for water)

a. 0.1M C₆H₁₂O₆

b. 0.1M CaCl₂

c. 0.1M NaCl

Answer 1

Balanced equation:
Na₂CO₃ + 2 HCl ===> 2 NaCl + H₂O + CO₂

Number of mole (mmoles) hydrochloric acid used = 25.5 x 0.3 /1000 =  0.00765 Moles

number of moles (mmoles) of sodium carbonate reacted =  0.003825 Moles

mass (grams) of sodium cobonate reacted =   0.003825 x 105.988 =  0.4054 gm

Balanced equation:
2 Al(OH)₃ + 3 H₂SO₄ ====> Al₂(SO₄)₃ + 6 H₂O
Reaction type: double replacement

number of mole (mmoles) sulfuric acid used = 75.5 x 0.5 /1000 =  0.03775 Moles

number of moles (mmoles) Aluminum oxide reacted =  0.02516 Moles

mass (grams) of aluminum hydroxide reacted =  0.01258 x 342.15 =  4.305 gm