OMPT chemistry practice: lesson and 10 problems

Answer: 74.1 g/mol

The subscript 2 sits outside the bracket, so it multiplies everything inside: one calcium, two oxygens, two hydrogens. That gives $40.1+2(16.0)+2(1.0)=40.1+32.0+2.0=74.1\text{g/mol}$. Reading the bracket as one $\mathrm{OH}$ group of mass $17.0$ and doubling it gives the same thing: $40.1+2(17.0)=74.1$.

Answer: 0.300 mol

The mole is a counting unit, and $n=\frac{m}{M}$ converts a mass into that count: $n=\frac{12.0}{40.0}=0.300\text{mol}$. Checking the units is the fastest way to know you divided the right way round: grams divided by grams per mole leaves moles.

Answer: B — $0.50\text{mol/L}$

Concentration is moles per litre, so the volume has to be in litres first: $500\text{mL}=0.500\text{L}$. Then $c=\frac{n}{V}=\frac{0.25}{0.500}=0.50\text{mol/L}$. Halving the volume of a solution doubles its concentration, and here the volume is half a litre, so the concentration is twice the number of moles.

Answer: 5

Leave oxygen until last, because it appears in two products. One ${\mathrm{C}}_3{\mathrm{H}}_8$ carries 3 carbons, so 3 $\mathrm{C}{\mathrm{O}}_2$; it carries 8 hydrogens, so 4 ${\mathrm{H}}_2\mathrm{O}$. Now count the oxygen on the right: $3\times 2+4\times 1=10$ atoms, which is 5 molecules of ${\mathrm{O}}_2$. The balanced equation is ${\mathrm{C}}_3{\mathrm{H}}_8+5{\mathrm{O}}_2\to 3\mathrm{C}{\mathrm{O}}_2+4{\mathrm{H}}_2\mathrm{O}$.

Answer: 7.20 g

Mass problems always travel through moles. First $n({\mathrm{C}}_3{\mathrm{H}}_8)=\frac{4.40}{44.0}=0.100\text{mol}$. The equation gives 4 moles of water per mole of propane, so $n({\mathrm{H}}_2\mathrm{O})=0.400\text{mol}$. Back to mass: $m=0.400\times 18.0=7.20\text{g}$.

Answer: C — $4.0\text{mol}$

Test each reactant against the ratio. The $4.0\text{mol}$ of ${\mathrm{H}}_2$ requires $2.0\text{mol}$ of ${\mathrm{O}}_2$, and $3.0\text{mol}$ is available, so oxygen is in excess and hydrogen runs out first. Hydrogen is the limiting reactant, and the $2:2$ ratio gives $4.0\text{mol}$ of water. A mole of ${\mathrm{O}}_2$ is left over untouched.

Answer: 2

Concentration first: $c=\frac{0.0025}{0.250}=0.010\text{mol/L}$. A strong acid dissociates completely, so $[{\mathrm{H}}^{+}]=0.010=10^{-2}\text{mol/L}$. Then $\mathrm{pH}=-{\log }_{10}[{\mathrm{H}}^{+}]=-{\log }_{10}(10^{-2})=2.00$. When the concentration is a clean power of ten, the pH is just the exponent with its sign flipped, and no calculator is needed.

Answer: 37.5 mL

Start from the acid: $n({\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4)=0.0250\times 0.15=0.00375\text{mol}$. Sulfuric acid is diprotic, and the equation says each mole needs two moles of base, so $n(\mathrm{NaOH})=0.00750\text{mol}$. Volume from concentration: $V=\frac{n}{c}=\frac{0.00750}{0.20}=0.0375\text{L}=37.5\text{mL}$.

Answer: Step 2

Step 1 is correct: the oxidation number of zinc rises, which is the definition of oxidation. Step 2 inverts the naming. The species that is oxidised hands its electrons to something else, so it is the reducing agent; zinc is the reductant here. The oxidising agent is the one that takes those electrons and is itself reduced, namely $\mathrm{C}{\mathrm{u}}^{2+}$, whose oxidation number falls from $+2$ to 0. Step 3 is the same inversion repeated, but Step 2 is where it first appears.

Answer: Step 2

The count in Step 1 is right. Step 2 breaks the one rule of balancing: you may change coefficients, never subscripts. ${\mathrm{H}}_2{\mathrm{O}}_2$ is hydrogen peroxide, a different substance from water, so Step 2 has balanced a reaction nobody asked about. Step 3 is arithmetically true and chemically irrelevant. The correct balance uses coefficients only: $2{\mathrm{H}}_2+{\mathrm{O}}_2\to 2{\mathrm{H}}_2\mathrm{O}$, giving 4 H and 2 O on each side.