Stoichiometry and Reactions Practice Questions & Answers

Avogadro's number is the number of constituent particles (usually atoms or molecules) that are contained in one mole of a given substance, defined as exactly $6.02214076 \times 10^{23}$6.02214076×1023.

To calculate moles: $\text{Moles} = \frac{\text{Atoms}}{\text{Avogadro's Number}} = \frac{2.50 \times 10^{24}}{6.022 \times 10^{23}} \approx 4.15 \text{ mol}$Moles=Avogadro’s NumberAtoms​=6.022×10232.50×1024​≈4.15 mol.

$\text{Mass} = \text{Moles} \times \text{Molar Mass} = 0.500 \times 40.08 = 20.04 \text{ g}$Mass=Moles×Molar Mass=0.500×40.08=20.04 g

1 mole of Ar contains $6.022 \times 10^{23}$6.022×1023 atoms. 1 mole of H$_2$2​ contains $6.022 \times 10^{23}$6.022×1023 molecules, but since each molecule has 2 atoms, it contains $1.2 \times 10^{24}$1.2×1024 atoms. This is greater than the other options.

First find moles: $\frac{36.0 \text{ g}}{18.02 \text{ g/mol}} \approx 2.0 \text{ mol}$18.02 g/mol36.0 g​≈2.0 mol. Then multiply by Avogadro's number: $2.0 \times 6.022 \times 10^{23} = 1.204 \times 10^{24}$2.0×6.022×1023=1.204×1024 molecules.

$3.01 \times 10^{23}$3.01×1023 is exactly 0.5 moles. Mass = $0.5 \text{ mol} \times 44.01 \text{ g/mol} = 22.0 \text{ g}$0.5 mol×44.01 g/mol=22.0 g.

Add the atomic masses: $22.99 + 35.45 = 58.44 \text{ g/mol}$22.99+35.45=58.44 g/mol.

Calculations: $1(\text{Ca}) + 2(\text{N}) + 6(\text{O})$1(Ca)+2(N)+6(O). $40.08 + 2(14.01) + 6(16.00) = 40.08 + 28.02 + 96.00 = 164.10 \text{ g/mol}$40.08+2(14.01)+6(16.00)=40.08+28.02+96.00=164.10 g/mol.

Molar mass is defined as the mass in grams of one mole of a substance, expressed in g/mol.

$6(12.01) + 12(1.008) + 6(16.00) = 72.06 + 12.096 + 96.00 \approx 180.16 \text{ g/mol}$6(12.01)+12(1.008)+6(16.00)=72.06+12.096+96.00≈180.16 g/mol.