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Answer:
0.26g of NaCl is the maximum mass that could be produced
Explanation:
Based on the reaction:
HCl + NaOH → NaCl + H₂O
<em>Where 1 mol of HCl reacts per mol of NaOH to produce 1 mol of NaCl</em>
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To solve this question we need to find <em>limiting reactant. </em>The moles of limiting reactant = Moles of NaCl produced:
<em>Moles HCl -Molar mass: 36.46g/mol-:</em>
0.365g HCl * (1mol / 36.46g) = 0.010 moles HCl
<em>Moles NaOH -Molar mass: 40g/mol-:</em>
0.18g NaOH * (1mol / 40g) = 0.0045 moles NaOH
As the reaction is 1:1 and moles NaOH < moles HCl, limiting reactant is NaOH and maximum moles produced of NaCl are 0.0045 moles.
The mass of NaCl is:
<em>Mass NaCl -Molar mass: 58.44g/mol-:</em>
0.0045 moles * (58.44g/mol) =
<h3>0.26g of NaCl is the maximum mass that could be produced</h3>
The molar mass of Fe would be 55.8450.
Answer:
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
Explanation:
Main reaction: 2Ag⁺(aq) + Mn(s) ⇄ 2Ag(s) + Mn²⁺(aq)
In the oxidation half reaction, the oxidation number increases:
Mn changes from 0, in the ground state to Mn²⁺.
The reduction half reaction occurs where the element decrease the oxidation number, because it is gaining electrons.
Silver changes from Ag⁺ to Ag.
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
To balance the hole reaction, we need to multiply by 2, the second half reaction:
Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
(Ag⁺(aq) + 1e⁻ ⇄ Ag(s)) . 2
2Ag⁺(aq) + 2e⁻ ⇄ 2Ag(s)
Now we sum, and we can cancel the electrons:
2Ag⁺(aq) + Mn(s) + 2e⁻ ⇄ 2Ag(s) + Mn²⁺(aq) + 2e⁻