Mole Concept and Stoichiometry Questions and Answers

Answer: B

The molar mass of nitric acid, HNO₃, is 63 g/mol.

Add the atomic masses of one hydrogen, one nitrogen and three oxygen atoms:

\[ \begin{aligned} M &= 1 + 14 + 3 \times 16 \\ &= 1 + 14 + 48 = 63\ \text{g/mol} \end{aligned} \]

48 g/mol is the mass of the three oxygen atoms alone. 62 g/mol is the nitrate ion, NO₃^-, without the hydrogen. 49 g/mol is the equivalent weight of sulphuric acid (98 divided by 2), a figure from a different calculation.

Answer: D

The maximum number of molecules is in 1.0 mol of O₂ gas.

Compare the amounts in moles, taking 22.4 L per mole at STP:

\[ \begin{aligned} 15\ \mathrm{L\ H_{2}} &= 15/22.4 = 0.67\ \mathrm{mol} \\ 10\ \mathrm{L\ O_{2}} &= 10/22.4 = 0.45\ \mathrm{mol} \\ 0.5\ \mathrm{g\ H_{2}} &= 0.5/2 = 0.25\ \mathrm{mol} \\ \mathrm{O_{2}} &= 1.0\ \mathrm{mol} \end{aligned} \]

The number of molecules is proportional to moles, so 1.0 mol, 6.02 × 10²³ molecules, is the largest. Hydrogen's low molar mass does not help when only 0.5 g is taken.

Answer: A

The E factor is 2.0.

The E factor is the mass of waste produced per unit mass of product. With 100% conversion, the only waste is CuI₂ (as the equation is written), and the product is biphenyl:

\[ \begin{aligned} M(\mathrm{C_{12}H_{10}}) &= 154 \\ M(\mathrm{CuI_{2}}) &= 63.5 + 2(126.9) = 317.3 \\ E &= \frac{317.3}{154} = 2.06 \approx 2.0 \end{aligned} \]

Mass balance checks it: 2 PhI (408) + Cu (63.5) = 471.5 = 154 + 317.3. An E factor of 1.0 would need equal masses of waste and product, and 4.0 or 8.3 overstate the waste.

Answer: C

There are \(14.69 \times 10^{23}\) bromine atoms.

One mole of any species holds Avogadro's number of particles, \(6.022 \times 10^{23}\). Here the amount is given as mol Br, that is moles of bromine atoms.

\[ \begin{aligned} N &= n \times N_{A} \\ &= 2.44 \times 6.022 \times 10^{23} \\ &= 14.69 \times 10^{23} \end{aligned} \]

The same number is \(1.469 \times 10^{24}\). The \(10^{22}\) option is ten times too small, \(146.9 \times 10^{24}\) is a hundred times too large, and \(1.46 \times 10^{23}\) is about ten times too small.

Answer: C

The mass is 14.3 g.

Mass is density times volume:

\[ \begin{aligned} m &= \rho V \\ &= 1.43\ \text{g/L} \times 10\ \text{L} \\ &= 14.3\ \text{g} \end{aligned} \]

1.43 g is the mass of one litre only, and 10 g simply copies the volume as a mass. 1 g has no basis in the data. As a check, oxygen's density of 1.43 g/L at STP is its molar mass 32 g divided by the molar volume 22.4 L.

Answer: C

The answer has 3 significant figures.

\[ 64.6312 - 64.5899 = 0.0413 \]

In addition and subtraction the result keeps as many decimal places as the least precise number; both have four decimal places, so the answer is 0.0413. Leading zeros only fix the decimal point and are not significant, which leaves 4, 1 and 3: three significant figures.

Six is the count in each starting number, and 4 counts the zero after the decimal point. Subtracting two nearly equal numbers loses significant figures, which is the point of the question.