34. 2NaHCO3→Na2CO3 + CO2 + H2O ushbu reaksiya bo'yicha 50% qo'shimchasi bo'lgan 168 g NaHCO3 sarflangan bo'lsa, necha mol H2O hamda necha litr (n.sh) CO2 hosil bo'lgan?

Kimyoviy reaksiya:

\( 2\text{NaHCO}_3 \rightarrow \text{Na}_2\text{CO}_3 + \text{CO}_2 \uparrow + \text{H}_2\text{O} \)

Berilgan:

\( m(\text{NaHCO}_3 \text{ (toza)}= 168 \text{ g} \times 50\% = 84 \text{ g} \)

Topish kerak:

\( n(\text{H}_2\text{O}) - ? \text{ mol} \)

\( V(\text{CO}_2 \text{ (n.sh)}) - ? \text{ litr} \)

Yechish:

1. Natriy gidrokarbonatning molyar massasini hisoblaymiz:

\( M(\text{NaHCO}_3) = 23 + 1 + 12 + 3 \cdot 16 = 84 \text{ g/mol} \)

2. Reaksiya natijasida sarflangan toza natriy gidrokarbonatning mol miqdorini aniqlaymiz:

\( n(\text{NaHCO}_3) = \frac{m(\text{NaHCO}_3)}{M(\text{NaHCO}_3)} = \frac{84 \text{ g}}{84 \text{ g/mol}} = 1 \text{ mol} \)

3. Reaksiya tenglamasiga ko'ra, 2 mol \( \text{NaHCO}_3 \) dan 1 mol \( \text{H}_2\text{O} \) hosil bo'ladi. Shu sababli, 1 mol \( \text{NaHCO}_3 \) dan hosil bo'ladigan \( \text{H}_2\text{O} \) molini topamiz:

\( n(\text{H}_2\text{O}) = \frac{n(\text{NaHCO}_3)}{2} = \frac{1 \text{ mol}}{2} = 0.5 \text{ mol} \)

4. Reaksiya tenglamasiga ko'ra, 2 mol \( \text{NaHCO}_3 \) dan 1 mol \( \text{CO}_2 \) hosil bo'ladi. Shu sababli, 1 mol \( \text{NaHCO}_3 \) dan hosil bo'ladigan \( \text{CO}_2 \) molini topamiz:

\( n(\text{CO}_2) = \frac{n(\text{NaHCO}_3)}{2} = \frac{1 \text{ mol}}{2} = 0.5 \text{ mol} \)

5. Normal sharoitda \( \text{CO}_2 \) hajmini hisoblaymiz (n.sh.da har qanday gazning 1 mol hajmi 22.4 litrga teng):

\( V(\text{CO}_2) = n(\text{CO}_2) \times V_m = 0.5 \text{ mol} \times 22.4 \text{ L/mol} = 11.2 \text{ L} \)

Ответ: 0,5 мол H2O ва 11,2 литр CO2 ҳосил бўлган.