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Chapter 22: Problem 72

Complete and balance the following equations: (a) \(\mathrm{CO}_{2}(g)+\mathrm{OH}^{-}(a q) \longrightarrow\) (b) \(\mathrm{NaHCO}_{3}(s)+\mathrm{H}^{+}(a q) \longrightarrow\) (c) \(\mathrm{CaO}(s)+\mathrm{C}(s) \stackrel{\Delta}{\longrightarrow}\) (d) \(\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \stackrel{\Delta}{\longrightarrow}\) (e) \(\mathrm{CuO}(s)+\mathrm{CO}(g) \longrightarrow\)

Short Answer

Expert verified
(a) CO鈧(g) + OH鈦(aq) 鈫 HCO鈧冣伝(aq) (b) NaHCO鈧(s) + H鈦(aq) 鈫 Na鈦(aq) + H鈧侰O鈧(aq) (c) CaO(s) + C(s) 鈫 CaC鈧(s) (d) C(s) + H鈧侽(g) 鈫 CO(g) + H鈧(g) (e) CuO(s) + CO(g) 鈫 Cu(s) + CO鈧(g)

Step by step solution

01

(a) Completing the equation

For the first equation, we need to figure out the product when CO鈧(g) reacts with OH鈦(aq). The product of this reaction should be a combination of the elements involved - carbon, oxygen, and hydrogen. The likely product would be HCO鈧冣伝(aq), which is a known compound and forms from these reactants. Now let's balance the equation.
02

(a) Balancing the equation

The unbalanced equation looks like this: CO鈧(g) + OH鈦(aq) 鈫 HCO鈧冣伝(aq) As we can see, each atom is already balanced on both sides of the equation. Therefore, the balanced equation is: CO鈧(g) + OH鈦(aq) 鈫 HCO鈧冣伝(aq)
03

(b) Completing the equation

For the second equation, the reactants are NaHCO鈧(s) and H鈦(aq). The likely product of this reaction is a combination of the elements involved, leading to the formation of sodium ions and carbonic acid: NaHCO鈧(s) + H鈦(aq) 鈫 Na鈦(aq) + H鈧侰O鈧(aq)
04

(b) Balancing the equation

The given equation is already balanced, so the final balanced equation is: NaHCO鈧(s) + H鈦(aq) 鈫 Na鈦(aq) + H鈧侰O鈧(aq)
05

(c) Completing the equation

In the third equation, CaO(s) and C(s) react in the presence of heat (螖). The product of this reaction should be calcium carbide (CaC鈧), which is formed by reacting calcium oxide with carbon at high temperatures: CaO(s) + C(s) 鈫 CaC鈧(s)
06

(c) Balancing the equation

The equation is already balanced, so the final balanced equation is: CaO(s) + C(s) 鈫 CaC鈧(s)
07

(d) Completing the equation

In the fourth equation, carbon (C) reacts with water (H鈧侽) in the presence of heat (螖). The products of this reaction are carbon monoxide (CO) and hydrogen gas (H鈧): C(s) + H鈧侽(g) 鈫 CO(g) + H鈧(g)
08

(d) Balancing the equation

Balancing the equation gives: C(s) + H鈧侽(g) 鈫 CO(g) + H鈧(g) This equation is already balanced.
09

(e) Completing the equation

In the last equation, copper oxide (CuO) reacts with carbon monoxide (CO). The products of this reaction are copper (Cu) and carbon dioxide (CO鈧): CuO(s) + CO(g) 鈫 Cu(s) + CO鈧(g)
10

(e) Balancing the equation

The given equation is already balanced, so the final balanced equation is: CuO(s) + CO(g) 鈫 Cu(s) + CO鈧(g)

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Most popular questions from this chapter

Complete and balance the following equations: (a) \(\mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) (b) \(\mathrm{Al}_{2} \mathrm{O}_{3}(s)+\mathrm{H}^{+}(a q) \longrightarrow\) (c) \(\mathrm{Na}_{2} \mathrm{O}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(I) \longrightarrow\) (d) \(\mathrm{N}_{2} \mathrm{O}_{3}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) (e) \(\mathrm{KO}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) (f) \(\mathrm{NO}(g)+\mathrm{O}_{3}(g) \longrightarrow\)

Write balanced equations for each of the following reactions (some of these are analogous to reactions shown in the chapter). (a) Aluminum metal reacts with acids to form hydrogen gas. (b) Steam reacts with magnesium metal to give magnesium oxide and hydrogen. (c) Manganese(IV) oxide is reduced to manganese(II) oxide by hydrogen gas. (d) Calcium hydride reacts with water to generate hydrogen gas.

Why are the properties of hydrogen different from those of either the group \(1 \mathrm{~A}\) or 7 A elements?

(a) Give the names and chemical symbols for the three isotopes of hydrogen. (b) List the isotopes in order of decreasing natural abundance.

In aqueous solution, hydrogen sulfide reduces (a) \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+}\), (b) \(\mathrm{Br}_{2}\) to \(\mathrm{Br}^{-}\), (c) \(\mathrm{MnO}_{4}^{-}\) to \(\mathrm{Mn}^{2+}\), (d) \(\mathrm{HNO}_{3}\) to \(\mathrm{NO}_{2}\). In all cases, under appropriate conditions, the product is elemental sulfur. Write a balanced net ionic equation for each reaction.
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