2.7 · Chemical ReactionsPractice · 化学反应

Work per 1 $\ce{CH3OH}$: balance C to get 1 $\ce{CO2}$; balance H, 4 H atoms give 2 $\ce{H2O}$. The right side has 2 + 2 = 4 O atoms, and the methanol already brings 1, so 3 more are needed: 3/2 $\ce{O2}$. Multiply everything by 2 to clear the fraction: $\ce{2CH3OH(l) +}$ $\ce{3O2(g) ->}$ $\ce{2CO2(g) +}$ $\ce{4H2O(l)}$. Don’t forget the oxygen atom already in the fuel.

Activity series (only the metals used here): Al > Zn > Fe > H > Cu > Ag. A metal displaces any metal below it from a solution of its salt, and metals above H displace hydrogen from acids.

zinc in $\ce{CuSO4(aq)}$: $\ce{Zn(s) +}$ $\ce{CuSO4(aq) ->}$ $\ce{ZnSO4(aq) +}$ $\ce{Cu(s)}$

Cu is below Fe, so copper in $\ce{FeSO4(aq)}$ does not react.

iron in dilute hydrochloric acid: $\ce{Fe(s) +}$ $\ce{2HCl(aq) ->}$ $\ce{FeCl2(aq) +}$ $\ce{H2(g)}$

Ag is below H, so silver in dilute hydrochloric acid does not react.

aluminum in $\ce{ZnSO4(aq)}$: $\ce{2Al(s) +}$ $\ce{3ZnSO4(aq) ->}$ $\ce{Al2(SO4)3(aq) +}$ $\ce{3Zn(s)}$

K, Ca, and Na are active enough to react with cold water: $\ce{Ca(s) +}$ $\ce{2H2O(l) ->}$ $\ce{Ca(OH)2(aq) +}$ $\ce{H2(g)}$. Magnesium barely reacts with cold water; it reacts with hot water: $\ce{Mg(s) +}$ $\ce{2H2O(l) ->[\Delta]}$ $\ce{Mg(OH)2(s) +}$ $\ce{H2(g)}$. Iron and copper do not react with water.

$\ce{MgCl2}$ + $\ce{KOH}$: swapping partners gives $\ce{Mg(OH)2}$ and $\ce{KCl}$; $\ce{Mg(OH)2}$ is insoluble and forms a white precipitate.

$\ce{Na2CO3}$ + $\ce{KCl}$: swapping partners gives $\ce{NaCl}$ and $\ce{K2CO3}$, both soluble, so no precipitate.

$\ce{BaCl2}$ + $\ce{K2CO3}$: swapping partners gives $\ce{BaCO3}$ and $\ce{KCl}$; $\ce{BaCO3}$ is insoluble and forms a white precipitate.

$\ce{CuSO4}$ + $\ce{NaNO3}$: swapping partners gives $\ce{Cu(NO3)2}$ and $\ce{Na2SO4}$, both soluble, so no precipitate.

$\ce{AgNO3}$ + $\ce{CaCl2}$: swapping partners gives $\ce{AgCl}$ and $\ce{Ca(NO3)2}$; $\ce{AgCl}$ is insoluble and forms a white precipitate.

All four ions form insoluble hydroxides with $\ce{OH-}$, but the colors differ: $\ce{Fe(OH)3}$ is reddish-brown, $\ce{Cu(OH)2}$ is blue, $\ce{Mg(OH)2}$ is white, $\ce{Al(OH)3}$ is white, gelatinous. A blue precipitate is $\ce{Cu(OH)2}$, so the solution contains $\ce{Cu^2+}$.

$\ce{Mg(OH)2}$ is insoluble, so it stays as a formula; HCl is a strong acid, fully ionized into $\ce{H+}$ and $\ce{Cl-}$, and $\ce{Cl-}$ is a spectator. The correct equation is $\ce{Mg(OH)2(s) +}$ $\ce{2H+(aq) ->}$ $\ce{Mg^2+(aq) +}$ $\ce{2H2O(l)}$. Writing $\ce{H+(aq) +}$ $\ce{OH-(aq) ->}$ $\ce{H2O(l)}$ wrongly splits up the insoluble $\ce{Mg(OH)2}$; $\ce{Mg(OH)2(s) +}$ $\ce{H+(aq) ->}$ $\ce{Mg^2+(aq) +}$ $\ce{H2O(l)}$ balances neither atoms nor charge; the last choice leaves the strong acid HCl unionized.

Only substances that dissolve and ionize completely are split: the strong acid $\ce{HNO3}$ and the soluble salt $\ce{Na2SO4}$. Insoluble $\ce{Cu(OH)2}$ and $\ce{AgCl}$ barely dissolve, and the weak acid $\ce{CH3COOH}$ ionizes only slightly, so all three stay as formulas.

$\ce{Fe2(SO4)3(s) ->}$ $\ce{2Fe^3+(aq) +}$ $\ce{3SO4^2-(aq)}$: each mole gives 3 mol of $\ce{SO4^2-}$, so 0.20 mol × 3 = 0.60 mol.

$\ce{Na2SO4(s) ->}$ $\ce{2Na+(aq) +}$ $\ce{SO4^2-(aq)}$: each mole gives 1 mol of $\ce{SO4^2-}$, so 0.10 mol × 1 = 0.10 mol.

Total: 0.60 + 0.10 = 0.70 mol. Note that one $\ce{Fe2(SO4)3}$ contains three $\ce{SO4^2-}$ ions.

$\ce{H2SO4}$ (a strong acid) and $\ce{MgCl2}$ (a soluble ionic compound) ionize completely: strong electrolytes. $\ce{HF}$ is a weak acid that ionizes only partly: a weak electrolyte, but still an electrolyte. Glucose dissolves as whole molecules: a nonelectrolyte. Graphite conducts with moving electrons, not ions, and gives no ions in water, so it is not an electrolyte.

In a decomposition, one reactant breaks into two or more products: heating $\ce{Cu(OH)2}$ gives CuO and water. In order: A synthesis, B decomposition, C single replacement, D double replacement.