What volume of oxygen is needed for the complete combustion of 5 g of magnesium to form magnesium oxide?

Molar mass of $\text{Mg} = 24.31 \text{ g/mol}$

$$\text{Moles of Mg} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{5.0 \text{ g}}{24.31 \text{ g/mol}} = 0.20567 \text{ mol}$$

From the balanced equation, $2 \text{ mol Mg}$ reacts with $1 \text{ mol O}_2$.

$$\text{Moles of O}_2 = \frac{1}{2} \times \text{Moles of Mg} = \frac{1}{2} \times 0.20567 \text{ mol} = 0.102835 \text{ mol}$$

At STP, $1 \text{ mol}$ of any gas occupies $22.4 \text{ dm}^3$.

$$\text{Volume of O}_2 = \text{Moles of O}_2 \times \text{Molar volume at STP}$$ $$\text{Volume of O}_2 = 0.102835 \text{ mol} \times 22.4 \text{ dm}^3/\text{mol} = 2.3035 \text{ dm}^3$$

Rounding to two significant figures:

$$\text{Volume of O}_2 = \boxed{\text{2.3 dm}^3}$$

: What volume of carbon dioxide is produced if Annlize uses 50 ml of 0,2 mol·dm⁻³ acetic acid?

The given reaction is:

$$\text{NaHCO}_3\text{(s)} + \text{CH}_3\text{COOH (aq)} \rightarrow \text{CH}_3\text{COONa (aq)} + \text{H}_2\text{O (l)} + \text{CO}_2\text{(g)}$$

$$\text{Moles of CH}_3\text{COOH} = \text{Concentration} \times \text{Volume}$$ $$\text{Moles of CH}_3\text{COOH} = 0.2 \text{ mol/dm}^3 \times 0.050 \text{ dm}^3 = 0.010 \text{ mol}$$

From the balanced equation, $1 \text{ mol CH}_3\text{COOH}$ produces $1 \text{ mol CO}_2$.

$$\text{Moles of CO}_2 = \text{Moles of CH}_3\text{COOH} = 0.010 \text{ mol}$$

At STP, $1 \text{ mol}$ of any gas occupies $22.4 \text{ dm}^3$.

$$\text{Volume of CO}_2 = \text{Moles of CO}_2 \times \text{Molar volume at STP}$$ $$\text{Volume of CO}_2 = 0.010 \text{ mol} \times 22.4 \text{ dm}^3/\text{mol} = 0.224 \text{ dm}^3$$

Rounding to two significant figures:

$$\text{Volume of CO}_2 = \boxed{\text{0.22 dm}^3}$$