Answer:
2.82 g
Explanation:
Step 1: Write the balanced precipitation reaction
3 Ba(NO₃)₂ (aq) + Al₂(SO₄)₃ (aq) ⇒ 3 BaSO₄(s) + 2 Al(NO₃)₃(aq)
Step 2: Calculate the reacting moles of Ba(NO₃)₂
45.0 mL (0.0450 L) of 0.548 M Ba(NO₃)₂ react.
0.0450 L × 0.548 mol/L = 0.0247 mol
Step 3: Calculate the moles of Al₂(SO₄)₃ that react with 0.0247 moles of Ba(NO₃)₂
The molar ratio of Ba(NO₃)₂ to Al₂(SO₄)₃ is 3:1. The reacting moles of Al₂(SO₄)₃ are 1/3 × 0.0247 mol = 8.23 × 10⁻³ mol
Step 4: Calculate the mass corresponding to 8.23 × 10⁻³ moles of Al₂(SO₄)₃
The molar mass of Al₂(SO₄)₃ is 342.2 g/mol.
8.23 × 10⁻³ mol × 342.2 g/mol = 2.82 g
The reaction between NaOH and HCl is as follows
NaOH + HCl ---> NaCl + H₂O
for neutralisation, H⁺ ions react with an equivalent amount of OH⁻ ions.
Number of NaOH moles reacted = 0.270 M/1000 mL/L x 37 mL = 0.00999 mol
number of HCl moles reacted = 0.270 M/1000 mL x 27 mL = 0.00729 mol
HCl reacts with NaOH in 1:1 molar ratio
Number of excess NaOH moles remaining - 0.00999 - 0.00729 = 0.0027 mol
total volume of solution = 37 mL + 27 mL = 64 mL = 0.064 L
Since there's excess OH⁻ ions, we can calculate pOH value first
pOH = - log [OH⁻]
[OH⁻] = 0.0027 mol / 0.064 L = 0.042 mol/L
pOH = -log(0.042 M)
pOH = 1.37
by knowing pOH we can calculate pH using the following equation;
pH + pOH = 14
pH = 14 - 1.37
pH = 12.63
Water is a compound. A compound is made up of two or more elements. In the case of water, or H2O, there are 2 hydrogen atoms present and 1 oxygen atom. I hope this helps you on your assignment. Good luck!
