Stoichiometry - Relative masses of atoms and molecules (3)
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1.
Sodium (Na) reacts with bromine (Br2) to form sodium bromide (NaBr). The balanced equation is:
2Na(s) + Br2(l) -> 2NaBr(s)
If 0.6 g of sodium reacts completely, calculate the mass of sodium bromide produced.
Calculation:
Step 1: Calculate the number of moles of Na.
Molar mass of Na = 22.99 g/mol
Moles of Na = mass / molar mass = 0.6 g / 22.99 g/mol = 0.0264 mol
Step 2: Calculate the number of moles of NaBr produced.
From the balanced equation, the mole ratio of Na:NaBr is 2:2 (or 1:1).
Therefore, 0.0264 mol of Na produces 0.0264 mol of NaBr.
Step 3: Calculate the mass of NaBr produced.
Molar mass of NaBr = 22.99 + 79.90 = 102.89 g/mol
Mass of NaBr = moles x molar mass = 0.0264 mol x 102.89 g/mol = 2.71 g
Answer: The mass of sodium bromide produced is 2.71 g.
2.
When 2.0 g of potassium (K) reacts with chlorine (Cl2), potassium chloride (KCl) is formed. The balanced equation for the reaction is:
2K(s) + Cl2(g) -> 2KCl(s)
Calculate the mass of potassium chloride produced.
Calculation:
Step 1: Calculate the number of moles of K.
Molar mass of K = 39.10 g/mol
Moles of K = mass / molar mass = 2.0 g / 39.10 g/mol = 0.0515 mol
Step 2: Calculate the number of moles of KCl produced.
From the balanced equation, the mole ratio of K:KCl is 2:2 (or 1:1).
Therefore, 0.0515 mol of K produces 0.0515 mol of KCl.
Step 3: Calculate the mass of KCl produced.
Molar mass of KCl = 39.10 + 35.45 = 74.55 g/mol
Mass of KCl = moles x molar mass = 0.0515 mol x 74.55 g/mol = 3.83 g
Answer: The mass of potassium chloride produced is 3.83 g.
3.
An element X has two isotopes: 20X (relative abundance = 60%) and 22X (relative abundance = 40%). Calculate the relative atomic mass of element X. Show your working.
To calculate the relative atomic mass (Ar) of element X, we use the following formula:
Ar = (Mass of isotope 1 x Relative abundance of isotope 1) + (Mass of isotope 2 x Relative abundance of isotope 2)
In this case:
- Mass of isotope 1 (20X) = 20 amu
- Relative abundance of isotope 1 = 60% = 0.60
- Mass of isotope 2 (22X) = 22 amu
- Relative abundance of isotope 2 = 40% = 0.40
Therefore:
Ar = (20 x 0.60) + (22 x 0.40) = 12 + 8.8 = 20.8 amu
So, the relative atomic mass of element X is 20.8.