Question

1. what is the mass of 2.05 moles of gallium (iii) sulfide (ga₂s₃)?
2. how many molecules are in 368 g of potassium hydroxide (koh)?
3. calcium chloride is a versatile chemical with uses ranging from road surfacing to water purification to cheesemaking. calcium chloride has a chemical formula of...

elo 4.3: i can determine the percent composition of a substance.

Explanation:

Problem 8: Mass of 2.05 moles of Gallium(III) sulfide ($\ce{Ga_{2}S_{3}}$)

Step 1: Calculate molar mass of $\ce{Ga_{2}S_{3}}$

Molar mass of Ga = $69.723\ \text{g/mol}$, molar mass of S = $32.065\ \text{g/mol}$.
For $\ce{Ga_{2}S_{3}}$:
Molar mass = $2\times69.723 + 3\times32.065$
= $139.446 + 96.195$
= $235.641\ \text{g/mol}$

Step 2: Calculate mass from moles

Mass = moles × molar mass
= $2.05\ \text{mol} × 235.641\ \text{g/mol}$
≈ $2.05 × 235.641$
≈ $483.06\ \text{g}$ (or check the given calculation steps, likely rounding differences)

Step 1: Calculate molar mass of $\ce{KOH}$

Molar mass of K = $39.098\ \text{g/mol}$, O = $15.999\ \text{g/mol}$, H = $1.0079\ \text{g/mol}$.
Molar mass = $39.098 + 15.999 + 1.0079$
= $56.1049\ \text{g/mol}$

Step 2: Calculate moles of $\ce{KOH}$

Moles = $\frac{\text{mass}}{\text{molar mass}}$ = $\frac{368\ \text{g}}{56.1049\ \text{g/mol}}$ ≈ $6.56\ \text{mol}$ (approximate, or as per the handwritten step: $368\div56.105\approx6.56$)

Step 3: Calculate number of molecules

Number of molecules = moles × Avogadro's number ($N_A = 6.022×10^{23}\ \text{molecules/mol}$)
= $6.56\ \text{mol} × 6.022×10^{23}\ \text{molecules/mol}$
≈ $3.95×10^{24}\ \text{molecules}$ (or follow the handwritten steps: $368\div56.105\approx6.56$, then $6.56×6.022×10^{23}\approx3.95×10^{24}$)

Step 1: Molar mass of $\ce{CaCl_{2}}$

Molar mass of Ca = $40.078\ \text{g/mol}$, Cl = $35.453\ \text{g/mol}$.
Molar mass = $40.078 + 2×35.453$
= $40.078 + 70.906$
= $110.984\ \text{g/mol}$

Step 2: Percent composition of Ca

% Ca = $\frac{40.078}{110.984}×100$ ≈ $36.11\%$

Step 3: Percent composition of Cl

% Cl = $\frac{70.906}{110.984}×100$ ≈ $63.89\%$ (or sum to 100%)

Answer:

Approximately $\boldsymbol{483\ \text{g}}$ (or as per precise calculation, ~483.06 g)

Problem 9: Molecules in 368 g of potassium hydroxide ($\ce{KOH}$)