If 100cm 3 of 0.05M hydrogen peroxide were used to produce gas M, calculate the volume of gas M produced if hydrogen peroxide was completely decomposed. (M.C.V=22.4 liters). 3mk

Mlinganyo wa mtengano wa haidrojeni peroksidi ni:

$$2\text{H}_2\text{O}_2\text{(aq)} \rightarrow 2\text{H}_2\text{O(l)} + \text{O}_2\text{(g)}$$

Gesi M ni oksijeni ($\text{O}_2$).

Badilisha ujazo wa haidrojeni peroksidi kutoka $\text{cm}^3$ hadi $\text{dm}^3$:

$$100 \text{ cm}^3 = \frac{100}{1000} \text{ dm}^3 = 0.100 \text{ dm}^3$$

Kokotoa moli za haidrojeni peroksidi:

$$\text{Moli za H}_2\text{O}_2 = \text{Mkusanyiko} \times \text{Ujazo (katika dm}^3)$$ $$\text{Moli za H}_2\text{O}_2 = 0.05 \text{ M} \times 0.100 \text{ dm}^3 = 0.005 \text{ moli}$$

Kutoka kwa mlinganyo, 2 moli za $\text{H}_2\text{O}_2$ hutoa 1 moli ya $\text{O}_2$.

$$\text{Moli za O}_2 = \frac{1}{2} \times \text{Moli za H}_2\text{O}_2$$ $$\text{Moli za O}_2 = \frac{1}{2} \times 0.005 \text{ moli} = 0.0025 \text{ moli}$$

Ujazo wa moli ya gesi (MGV) umetolewa kama $22.4 \text{ lita}$ (au $\text{dm}^3$).

$$\text{Ujazo wa O}_2 = \text{Moli za O}_2 \times \text{MGV}$$ $$\text{Ujazo wa O}_2 = 0.0025 \text{ moli} \times 22.4 \text{ dm}^3/\text{moli} = 0.056 \text{ dm}^3$$

Ujazo wa gesi M iliyozalishwa ni $\text{0.056 dm}^3$.