<u>A</u><u>)</u><u> </u><u>The moles of NaOH are equal to moles of HCl.</u>
Explanation:
NaOH + HCl ↦NaCl + H20
Answer:
22.96 $ per 1 g of gold
Explanation:
We know that 1 g is equal to 0.035 oz.
Now we formulate the following reasoning:
if 1 oz of gold have a value of 655 $
then 0.035 oz of gold have a value of X $
X = (0.035 × 655) / 1 = 22.96 $
Leftover: approximately 11.73 g of sulfuric acid.
<h3>Explanation</h3>
Which reactant is <em>in excess</em>?
The theoretical yield of water from Al(OH)₃ is lower than that from H₂SO₄. As a result,
- Al(OH)₃ is the limiting reactant.
- H₂SO₄ is in excess.
How many <em>moles</em> of H₂SO₄ is consumed?
Balanced equation:
2 Al(OH)₃ + 3 H₂SO₄ → Al₂(SO₄)₃ + 6 H₂O
Each mole of Al(OH)₃ corresponds to 3/2 moles of H₂SO4. The formula mass of Al(OH)₃ is 78.003 g/mol. There are 15 / 78.003 = 0.19230 moles of Al(OH)₃ in the five grams of Al(OH)₃ available. Al(OH)₃ is in excess, meaning that all 0.19230 moles will be consumed. Accordingly, 0.19230 × 3/2 = 0.28845 moles of H₂SO₄ will be consumed.
How many <em>grams</em> of H₂SO₄ is consumed?
The molar mass of H₂SO₄ is 98.076 g.mol. The mass of 0.28845 moles of H₂SO₄ is 0.28845 × 98.076 = 28.289 g.
How many <em>grams</em> of H₂SO₄ is in excess?
40 grams of sulfuric acid H₂SO₄ is available. 28.289 grams is consumed. The remaining 40 - 28.289 = 11.711 g is in excess. That's closest to the first option: 11.73 g of sulfuric acid.
The balanced net equation for
BaCl2 (aq) + H2SO4(aq) → BaSO4(s) + HCl (aq) is
Ba^2+(aq) +SO4^2- → BaSO4 (s)
<u><em>Explanation</em></u>
Ionic equation is a chemical equation in which electrolytes in aqueous solution are written as dissociated ions.
<u>ionic equation is written using the below steps</u>
Step 1: <em>write a balanced molecular equation</em>
BaCl2 (aq) +H2SO4 (aq)→ BaSO4(s) +2HCl (aq)
Step 2: <em>Break all soluble electrolytes in to ions</em>
= Ba^2+ (aq) + 2Cl^-(aq) + 2H^+(aq) + SO4^2-(aq)→ BaSO4(s) + 2H^+(aq) +2Cl^- (aq)
step 3: <em>cancel the spectator ions in both side of equation ( ions which do not take place in the reaction)</em>
<em> </em><em> =</em> 2Cl^- and 2H^+ ions
Step 4: <em>write the final net equation</em>
<em> Ba^2+(aq) + SO4^2-(aq)→ BaSO4(s</em><em>)</em>
Answer: the answer is C oxidizing
