How many sodium atoms are there in 0.3 moles of Na atoms? 2) A sample of carbon dioxide is made up of 1.21×10^23 molecules. Calculate how many moles of molecules the sample contains. 3) A bottle of cleaner “with ammonia” contains 1.7 g of NH3. Calculate how many molecules of ammonia are in the bottle.
1) How many sodium atoms are there in 0.3 moles of Na atoms?
2) A sample of carbon dioxide is made up of 1.21×10^23 molecules. Calculate how many moles of molecules the sample contains.
3) A bottle of cleaner “with ammonia” contains 1.7 g of NH3. Calculate how many molecules of ammonia are in the bottle.
4) The tank of a gas car contains 15 kg of methane, CH4. Determine how many moles of molecules make up that mass of substance.
The Correct Answer and Explanation is:
Let’s solve each question step-by-step and then provide a full explanation at the end.
1) How many sodium atoms are there in 0.3 moles of Na atoms?
We use Avogadro’s number, which is:
$$
6.022 \times 10^{23} \text{ atoms/mole}
$$
$$
0.3 \text{ mol Na} \times 6.022 \times 10^{23} \text{ atoms/mol} = \boxed{1.81 \times 10^{23} \text{ Na atoms}}
$$
2) A sample of carbon dioxide is made up of $1.21 \times 10^{23}$ molecules. Calculate how many moles of molecules the sample contains.
Use the inverse of Avogadro’s number:
$$
\frac{1.21 \times 10^{23}}{6.022 \times 10^{23}} = \boxed{0.201 \text{ moles CO₂}}
$$
3) A bottle of cleaner “with ammonia” contains 1.7 g of NH₃. Calculate how many molecules of ammonia are in the bottle.
Step 1: Find moles of NH₃.
Molar mass of NH₃ = 14.01 (N) + 3 × 1.008 (H) = 17.034 g/mol
$$
\frac{1.7 \text{ g}}{17.034 \text{ g/mol}} = 0.0998 \text{ mol NH₃}
$$
Step 2: Convert moles to molecules.
$$
0.0998 \text{ mol} \times 6.022 \times 10^{23} = \boxed{6.01 \times 10^{22} \text{ molecules NH₃}}
$$
4) The tank of a gas car contains 15 kg of methane, CH₄. Determine how many moles of molecules make up that mass of substance.
Step 1: Convert kg to grams.
$$
15 \text{ kg} = 15,000 \text{ g}
$$
Step 2: Find the molar mass of CH₄.
Carbon (C) = 12.01 g/mol
Hydrogen (H) = 1.008 g/mol × 4 = 4.032 g/mol
Total = 16.042 g/mol
$$
\frac{15,000 \text{ g}}{16.042 \text{ g/mol}} = \boxed{935.2 \text{ mol CH₄}}
$$
✨ 300-Word Explanation:
Understanding atoms and molecules in chemical quantities begins with Avogadro’s number, $6.022 \times 10^{23}$, which represents the number of particles (atoms, ions, or molecules) in one mole of any substance.
Problem 1 asks how many sodium atoms are in 0.3 moles. Since each mole contains $6.022 \times 10^{23}$ atoms, multiplying this value by 0.3 gives $1.81 \times 10^{23}$ atoms. This is a direct application of Avogadro’s number.
Problem 2 is the reverse. You are given the number of molecules and asked for the number of moles. Divide the number of particles by Avogadro’s number to find the number of moles. Here, $1.21 \times 10^{23}$ divided by $6.022 \times 10^{23}$ gives approximately 0.201 moles of CO₂.
Problem 3 combines molar mass and Avogadro’s number. First, convert the mass of ammonia (NH₃) into moles by dividing by its molar mass (17.034 g/mol). Once you know the moles (0.0998 mol), multiply by Avogadro’s number to get molecules, resulting in $6.01 \times 10^{22}$ molecules.
Problem 4 involves a large mass of methane (15 kg). Convert to grams (15,000 g), then divide by CH₄’s molar mass (16.042 g/mol) to get moles—approximately 935.2 mol. This is crucial in chemistry and real-world applications like calculating fuel needs or emissions.
These problems show how chemists move between mass, moles, and particles using consistent formulas and conversion factors, forming the foundation for chemical calculations in both laboratories and industry
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