Unit 4 · Chemical ReactionsPractice · 化学反应

The test is whether new substances form (atoms rearrange). In A, Zn and HCl become $\ce{ZnCl2}$ and $\ce{H2}$; in D, $\ce{H2O2}$ decomposes into $\ce{H2O}$ and $\ce{O2}$ ($\ce{MnO2}$ is a catalyst). B is sublimation and C is crystallization: the substances are still $\ce{I2}$ and NaCl. In E, intermolecular forces pack the molecules more tightly, but they’re still ethanol and water. A gas or a new color is only possible evidence; what matters is whether a new substance forms.

The soluble strong electrolytes $\ce{Na3PO4}$, $\ce{CaCl2}$, and NaCl all exist as ions in solution. The only real change is $\ce{Ca^2+}$ and $\ce{PO4^3-}$ combining into a $\ce{Ca3(PO4)2}$ precipitate; $\ce{Na+}$ and $\ce{Cl-}$ are spectators. The precipitate is neutral, 3 × (+2) + 2 × (−3) = 0, so the ions combine 3 : 2. A is the molecular equation, not the net ionic one. In B, NaCl is soluble and doesn’t precipitate. D’s charges don’t balance (2 − 3 = −1 on the left, 0 on the right), and its formula is wrong.

A net ionic equation splits strong electrolytes into ions, keeps weak electrolytes and precipitates as formulas, drops the spectator ions, and balances both atoms and charge. A: HF is a weak acid, so it stays as a molecule; correct. B: $\ce{HNO2}$ is also weak and can’t be written as $\ce{H+}$; it should be $\ce{HNO2(aq) +}$ $\ce{OH-(aq) ->}$ $\ce{NO2-(aq) +}$ $\ce{H2O(l)}$. C is correct. D keeps the spectator ions $\ce{NO3-}$ and $\ce{K+}$. E’s charges don’t balance (+1 on the left, +2 on the right); it should be $\ce{Mg(s) +}$ $\ce{2H+(aq) ->}$ $\ce{Mg^2+(aq) +}$ $\ce{H2(g)}$.

Count: before, 3 $\ce{N2}$ and 2 $\ce{O2}$; after, 4 NO and 1 leftover $\ce{N2}$. The leftover $\ce{N2}$ didn’t react, so it doesn’t belong in the equation. What reacted is 2 $\ce{N2}$ and 2 $\ce{O2}$, forming 4 NO; reduced to lowest whole numbers: $\ce{N2(g) +}$ $\ce{O2(g) ->}$ $\ce{2NO(g)}$. Check: 6 N and 4 O atoms both before and after. A includes the leftover $\ce{N2}$, C isn’t reduced, and D doesn’t balance the atoms.

A chemical change breaks bonds and rearranges atoms. When HCl dissolves, the covalent H–Cl bond (average bond enthalpy 431 kJ/mol) breaks, H⁺ moves to a water molecule, and a new O–H bond forms, giving new particles, $\ce{H3O+}$ and $\ce{Cl-}$. Sugar molecules stay whole when sugar dissolves; only intermolecular forces are overcome, which is different. No chemical reaction makes new elements, so C’s reason fails, and warming is only possible evidence, since dissolving alone can release heat, so D’s reason isn’t enough.

Convert to moles: n(Al) = 4.00 ÷ 26.98 = 0.148 mol; n($\ce{Cl2}$) = 10.0 ÷ 70.90 = 0.141 mol. 0.148 mol of Al would need 0.148 × 3/2 = 0.222 mol of $\ce{Cl2}$, more than is available, so $\ce{Cl2}$ is the limiting reactant. n($\ce{AlCl3}$) = 0.141 × 2/3 = 0.0940 mol, and m = 0.0940 × 133.33 = 12.5 g. Working from Al instead gives 19.8 g, more than the chlorine can supply.

The two members of a conjugate pair differ by exactly one $\ce{H+}$. $\ce{H2PO4-}$ gives an $\ce{H+}$ to $\ce{NH3}$ and becomes $\ce{HPO4^2-}$, so $\ce{H2PO4-}$ is the acid and $\ce{HPO4^2-}$ its conjugate base. $\ce{NH3}$ accepts the $\ce{H+}$ and becomes $\ce{NH4+}$, so $\ce{NH3}$ is a base and $\ce{NH4+}$ its conjugate acid. B pairs the two reactants (an acid and a base), C pairs two bases, and D pairs two acids.

$\ce{H2SO4(aq) +}$ $\ce{2NaOH(aq) ->}$ $\ce{Na2SO4(aq) +}$ $\ce{2H2O(l)}$: $\ce{H2SO4}$ and NaOH react 1 : 2.

M = 0.001570 mol ÷ 0.02500 L = 0.06280 M. All the data have 4 significant figures. Ignoring the 1 : 2 ratio gives 0.1256 M.

The leftover water dilutes the NaOH, so the same amount of acid needs a larger volume of base. The calculation still multiplies the labeled concentration by that volume, so the moles of NaOH, and the acid’s concentration, come out too high. If the NaOH is diluted to 0.95 of its labeled concentration, the result is about 5.3% too high. B is about the solution in the buret, but the calculation uses labeled concentration × volume used, and that product is too large.

A redox reaction is one in which some element changes oxidation number. B: Na goes from 0 to +1 and Cl from 0 to −1. D: O is −1 in $\ce{H2O2}$; some goes down to −2 (in $\ce{H2O}$) and some up to 0 (in $\ce{O2}$), so the same element is both oxidized and reduced. In A (decomposition), C (precipitation), and E (synthesis), no oxidation number changes, so they aren’t redox. Decomposition and synthesis reactions may or may not be redox.

The oxidation number of Cr in $\ce{Cr2O7^2-}$: 2x + 7 × (−2) = −2, so x = +6. Each Cr drops from +6 to +3, gaining 3 electrons, so the 2 Cr gain 6. Each $\ce{Fe^2+}$ loses 1 electron to become $\ce{Fe^3+}$, so the 6 lose 6. Electrons lost equal electrons gained: 6 are transferred. H and O don’t change oxidation number.

Mn goes from +4 (in $\ce{MnO2}$) to +2 (in $\ce{MnCl2}$), gaining 2 electrons: it is reduced, and $\ce{MnO2}$ is the oxidizing agent. Cl is −1 in HCl. The 2 that end up in $\ce{Cl2}$ rise to 0 and are oxidized, losing 2 electrons in all; the 2 in $\ce{MnCl2}$ stay at −1. So half the HCl acts as the reducing agent and half simply acts as an acid. H stays +1 throughout; the gas formed is $\ce{Cl2}$, not $\ce{H2}$.