To determine the purity of limestone, form 3 students heated 12.5g of limestone in a crucible until they obtained a constant mass. If the volume of CO they obtained was 240cm . Calculate the purity of limestone. (C = 12, O = 16, Ca = 40, Mav = 24.0dm )

Haya Robert, twende na swali hili la kemia.

Ili kukokotoa usafi wa chokaa, tunahitaji kwanza kujua ni kiasi gani cha kalsiamu kabonati ($\text{CaCO}_3$) safi kilichokuwepo kwenye sampuli ya chokaa. Chokaa hupungua kwa joto na kutoa kaboni dioksidi ($\text{CO}_2$).

• Hatua ya 1: Andika mlinganyo wa kemikali uliosawazishwa. Mlinganyo wa upungufu wa joto wa kalsiamu kabonati ni:

  $$\text{CaCO}_3(s) \to \text{CaO}(s) + \text{CO}_2(g)$$

  Kutoka kwenye mlinganyo, moli 1 ya $\text{CaCO}_3$ hutoa moli 1 ya $\text{CO}_2$.

• Hatua ya 2: Kokotoa idadi ya moli za $\text{CO}_2$ zilizozalishwa. Kiasi cha $\text{CO}_2$ kilichopatikana ni $240 \text{ cm}^3$. Tunakibadilisha kuwa $\text{dm}^3$:

  $$240 \text{ cm}^3 = \frac{240}{1000} \text{ dm}^3 = 0.240 \text{ dm}^3$$

  Kutumia kiasi cha moli cha gesi ($\text{Mav} = 24.0 \text{ dm}^{-3}$):

  $$\text{Moli za CO}_2 = \frac{\text{Kiasi cha CO}_2}{\text{Kiasi cha moli cha gesi}}$$ $$\text{Moli za CO}_2 = \frac{0.240 \text{ dm}^3}{24.0 \text{ dm}^{-3}} = 0.01 \text{ mol}$$

• Hatua ya 3: Kokotoa masi ya $\text{CaCO}_3$ safi iliyopungua. Kutoka Hatua ya 1, moli za $\text{CaCO}_3$ = moli za $\text{CO}_2$. Kwa hiyo, moli za $\text{CaCO}_3 = 0.01 \text{ mol}$. Kokotoa masi ya moli ya $\text{CaCO}_3$:

  $$\text{Masi ya moli ya CaCO}_3 = \text{Ca} + \text{C} + (3 \times \text{O})$$ $$\text{Masi ya moli ya CaCO}_3 = 40 + 12 + (3 \times 16) = 40 + 12 + 48 = 100 \text{ g/mol}$$

  Sasa, kokotoa masi ya $\text{CaCO}_3$ safi:

  $$\text{Masi ya CaCO}_3 = \text{Moli} \times \text{Masi ya moli}$$ $$\text{Masi ya CaCO}_3 = 0.01 \text{ mol} \times 100 \text{ g/mol} = 1.0 \text{ g}$$

• Hatua ya 4: Kokotoa asilimia ya usafi wa chokaa. Masi ya chokaa isiyo safi iliyopashwa moto ni $12.5 \text{ g}$.

  $$\text{Asilimia ya usafi} = \frac{\text{Masi ya CaCO}_3 \text{ safi}}{\text{Masi ya chokaa isiyo safi}} \times 100\%$$ $$\text{Asilimia ya usafi} = \frac{1.0 \text{ g}}{12.5 \text{ g}} \times 100\%$$ $$\text{Asilimia ya usafi} = 0.08 \times 100\% = 8\%$$

Asilimia ya usafi wa chokaa ni \text{8%}.