Module 6 · Chemical EquationsPractice · 化学方程式

In $\ce{2H2(g) +}$ $\ce{O2(g) ->}$ $\ce{2H2O(l)}$, $\ce{H2}$ and $\ce{O2}$ react in a 2 : 1 ratio. 4 $\ce{H2}$ molecules need only 2 $\ce{O2}$ molecules, forming 4 $\ce{H2O}$ molecules and leaving 1 $\ce{O2}$ molecule unreacted. The kinds and numbers of atoms don’t change: 8 H atoms and 6 O atoms before and after. A is missing 2 O atoms, and C has 4 extra H atoms. D has the right atoms, but the leftover oxygen stays as $\ce{O2}$ molecules; it doesn’t split into single O atoms.

The law of conservation of mass says the total mass of the substances that react equals the total mass of the products. Magnesium isn’t the only reactant: oxygen from the air reacts too, so the magnesium oxide is heavier than the magnesium by exactly the mass of oxygen that combined with it. Every chemical reaction obeys the law (not A), heat isn’t matter and can’t be weighed (not B), and atoms don’t change in a reaction; they only rearrange (not C).

First use conservation of mass to find how much D forms from 6 g of A: 6 g + 8 g − 9 g = 5 g. The mass ratio of the substances in a reaction is fixed, so A : D = 6 : 5, and 3 g of A forms 5 g × 3/6 = 2.5 g of D. The problem gives masses directly, so the coefficient 2 isn’t needed here.

Count the right side: $\ce{2CO2}$ and $\ce{4H2O}$ contain 2 C, 8 H, and 8 O atoms in all. On the left, $\ce{3O2}$ already supplies 6 O atoms. So 2 X together contain 2 C, 8 H, and 2 O atoms, and each X has 1 C, 4 H, and 1 O: $\ce{CH4O}$ (methanol, also written $\ce{CH3OH}$). $\ce{C2H8O2}$ forgets to divide by the coefficient 2 in front of X, and $\ce{CH4}$ leaves out the O in X.

The mass ratio is the ratio of coefficient × formula mass: (4 × 27) : (3 × 32) : (2 × 102) = 108 : 96 : 204 = 9 : 8 : 17. 4 : 3 : 2 is the ratio of particles; 27 : 32 : 102 forgets the coefficients; 9 : 4 : 17 uses 16 as the formula mass of $\ce{O2}$. Check: 108 + 96 = 204, as conservation of mass requires.

An equation can be read three ways. A reads the substances and the condition. B reads the particles: the coefficients give the ratio of molecules. D reads the masses, using coefficient × formula mass: 16 : (2 × 32) : 44 : (2 × 18) = 16 : 64 : 44 : 36, and 16 + 64 = 44 + 36. C mistakes the coefficient ratio for a mass ratio. E: there are 3 molecules before and 3 after. The number of molecules may or may not change; what never changes is the kinds and numbers of atoms.

Balance C and H first: $\ce{C2H6 + O2 -> 2CO2 + 3H2O}$. Now the right side has 2 × 2 + 3 = 7 O atoms, an odd number, but each $\ce{O2}$ brings 2 O atoms, so no whole number works. Double every coefficient: $\ce{2C2H6(g) +}$ $\ce{7O2(g) ->}$ $\ce{4CO2(g) +}$ $\ce{6H2O(l)}$. Check: C 4 = 4, H 12 = 12, O 14 = 14.

Each CO takes 1 O atom and becomes $\ce{CO2}$. $\ce{Fe3O4}$ has 4 O atoms, so it needs 4 CO, which form 4 $\ce{CO2}$. Then balance Fe: 3. $\ce{Fe3O4(s) +}$ $\ce{4CO(g) ->[\Delta]}$ $\ce{3Fe(s) +}$ $\ce{4CO2(g)}$. Check: Fe 3 = 3, C 4 = 4, O 4 + 4 = 8.

In a single-replacement reaction, an element reacts with a compound to form a different element and a different compound. In D, the element Fe reacts with the compound $\ce{H2SO4}$ to form the element $\ce{H2}$ and the compound $\ce{FeSO4}$. A produces the element Cu, but both reactants, CO and CuO, are compounds, so it isn’t single replacement. B is synthesis (many → one), and C is decomposition (one → many).

In a double-replacement reaction, two compounds swap partners to form two new compounds. A: NaOH and HCl swap partners to form NaCl and $\ce{H2O}$ (neutralization is a kind of double replacement). C: $\ce{BaCl2}$ and $\ce{Na2SO4}$ swap partners to form a $\ce{BaSO4}$ precipitate and NaCl. B has two compounds reacting, but they don’t swap partners, so it isn’t double replacement. D produces the element Fe even though no element reacts, so it isn’t any of the four basic types.

Let x be the mass of hydrogen peroxide.

$\ce{2H2O2(aq) ->[\ce{MnO2}]}$ $\ce{2H2O(l) +}$ $\ce{O2(g)}$

Write 2 × 34 = 68 under $\ce{2H2O2}$ and 32 under $\ce{O2}$. The proportion $\frac{68}{32} = \frac{x}{3.2\,\text{g}}$ gives x = 6.8 g, so at least 6.8 g of hydrogen peroxide is needed. $\ce{MnO2}$ is a catalyst: it isn’t used up, so it doesn’t enter the calculation.

By conservation of mass, the 4.4 g lost is the mass of $\ce{CO2}$ that escaped. The acid is in excess, so all the calcium carbonate reacts and the acid’s mass isn’t needed. Let x be the mass of calcium carbonate:

$\ce{CaCO3(s) +}$ $\ce{2HCl(aq) ->}$ $\ce{CaCl2(aq) +}$ $\ce{H2O(l) +}$ $\ce{CO2(g)}$

$\frac{100}{44} = \frac{x}{4.4\,\text{g}}$, so x = 10 g. Mass percent = 10 g ÷ 12.5 g × 100% = 80%. 35.2% treats the mass of $\ce{CO2}$ as the mass of calcium carbonate; 64.8% treats the remaining mass as calcium carbonate.