48.4 g C3H8 gas vapor water vol 1.00 g/mL

Calculate the volume of liquid water (density 1.00 g/mL)

produced by burning 48.4 g of propane and

condensing the gaseous water produced.

Assume lots of O2.

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프로판 연소 반응식

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

( 참고 https://ywpop.tistory.com/5791 )

C3H8의 몰질량 = 3(12) + 8(1) = 44 g/mol 이므로,

48.4 g × (1 mol / 44 g)

= 48.4 / 44

= 1.1 mol C3H8(g)

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

C3H8 : H2O = 1 : 4 = 1.1 mol : ? mol

? = 4 × 1.1 = 4.4 mol H2O(g)

H2O의 몰질량 = 18 g/mol 이므로,

(4.4 mol) × (18 g / 1 mol) × (1 mL / 1 g)

= 79.2 mL H2O(l)

(48.4 g C3H8) (1 mol C3H8 / 44 g C3H8) (4 mol H2O / 1 mol C3H8) (18 g H2O / 1 mol H2O) (1 mL H2O / 1 g H2O)

= (48.4) (1 / 44) (4 / 1) (18 / 1) (1 / 1)

= 79.2 mL H2O

답: 79.2 mL

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