Question

question 10
1 pts
which of the following is an oxidation - reduction reaction? (pick all that apply)
2 kclo₃(s) → 2 kcl(s) + 3 o₂(g)
nai(aq) + agno₃(aq) → agi(s) + nano₃(aq)
hcl(aq) + lioh(aq) → licl(aq) + h₂o(l)
c₆h₁₂o₂(l) + 8 o₂(g) → 6 co₂(g) + 6 h₂o(g)
mg(s) + 2 hcl(aq) → mgcl₂(aq) + h₂(g)
question 11
1 pts
the following reaction can be classified as what type(s) of reaction(s)?
2 al(oh)₃(aq) + 3 h₂so₄(aq) → al₂(so₄)₃(s) + 6 h₂o(l)
precipitation
acid - base
oxidation and reduction
precipitation and acid - base

Explanation:

Question 10

• For \(2KClO_{3}(s)\to2KCl(s) + 3O_{2}(g)\):
• In \(KClO_{3}\), \(Cl\) has an oxidation state of \(+5\) (\(K = + 1\), \(O=-2\), let \(Cl=x\), then \(+1+x+3\times(-2)=0\), \(x = + 5\)) and in \(KCl\), \(Cl\) has an oxidation state of \(-1\). \(O\) has an oxidation state of \(-2\) in \(KClO_{3}\) and \(0\) in \(O_{2}\). So, there is a change in oxidation states.
• For \(NaI(aq)+AgNO_{3}(aq)\to AgI(s)+NaNO_{3}(aq)\):
• This is a double - displacement reaction. Oxidation states of \(Na(+1)\), \(I(-1)\), \(Ag(+1)\), \(N(+5)\), \(O(-2)\) remain the same on both sides of the equation.
• For \(HCl(aq)+LiOH(aq)\to LiCl(aq)+H_{2}O(l)\):
• This is an acid - base (neutralization) reaction. Oxidation states of \(H(+1)\), \(Cl(-1)\), \(Li(+1)\), \(O(-2)\) remain the same.
• For \(C_{6}H_{12}O_{2}(l)+8O_{2}(g)\to6CO_{2}(g)+6H_{2}O(g)\):
• In \(C_{6}H_{12}O_{2}\), \(C\) has an average oxidation state (let \(C\) oxidation state be \(x\), \(6x + 12\times(+1)+2\times(-2)=0\), \(6x=-8\), \(x=-\frac{4}{3}\)). In \(CO_{2}\), \(C\) has an oxidation state of \(+4\). \(O\) has an oxidation state of \(0\) in \(O_{2}\) and \(-2\) in \(CO_{2}\) and \(H_{2}O\). So, there is a change in oxidation states.
• For \(Mg(s)+2HCl(aq)\to MgCl_{2}(aq)+H_{2}(g)\):
• \(Mg\) has an oxidation state of \(0\) (in \(Mg(s)\)) and \(+2\) (in \(MgCl_{2}\)). \(H\) has an oxidation state of \(+1\) (in \(HCl\)) and \(0\) (in \(H_{2}\)). So, there is a change in oxidation states.

• For \(2Al(OH)_{3}(aq)+3H_{2}SO_{4}(aq)\to Al_{2}(SO_{4})_{3}(s)+6H_{2}O(l)\):
• It is an acid - base reaction because \(Al(OH)_{3}\) (a base) reacts with \(H_{2}SO_{4}\) (an acid). Also, \(Al_{2}(SO_{4})_{3}\) is formed as a precipitate.

Answer:

\(2KClO_{3}(s)\to2KCl(s) + 3O_{2}(g)\), \(C_{6}H_{12}O_{2}(l)+8O_{2}(g)\to6CO_{2}(g)+6H_{2}O(g)\), \(Mg(s)+2HCl(aq)\to MgCl_{2}(aq)+H_{2}(g)\)

Question 11