25℃ 855 mmHg 190 g Al react O2 vol

25℃, 855 mmHg에서 190 g의 알루미늄과

반응하기 위해 필요한 산소의 부피(L)를 구하시오.

4Al(s) + 3O2(g) → 2Al2O3(s)

Oxygen gas reacts with powdered aluminum

according to the following reaction:

4Al(s) + 3O2(g) → 2Al2O3(s)

What volume of O2 gas (in L),

measured at 855 mmHg and 25℃,

is required to completely react with 190 g of Al?

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Al의 몰질량 = 26.98 g/mol 이므로,

190 g / (26.98 g/mol) = 7.04225 mol Al

4Al(s) + 3O2(g) → 2Al2O3(s)

Al : O2 = 4 : 3 = 7.04225 mol : ? mol

? = 3 × 7.04225 / 4 = 5.2816875 mol O2

PV = nRT 로부터,

5.2816875 mol O2의 부피를 계산하면,

V = nRT / P

= (5.2816875) (0.08206) (273.15 + 25) / (855/760)

= 114.86 L

답: 약 115 L

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