3. The thermochemical equation for the combustion of propane gas is as follows:

Question

3. The thermochemical equation for the combustion of propane gas is as follows: C3H8(g) + 502(g) 3CO2(g) + 4H20(g); 110-2046 kJ/mol C3H8 (a) How much heat is produced from the complete combustion of 1.00 g of propane (C3Hs)? (b) The specific heat of water is 4.184 J/g.°C. How much energy (in k.J) is needed to heat 1.00 qt of water from 20.0°C to 100.0°C (1 .00 qt-946.4 g) (c) How many grams of propane gas (molar mass- 44.06 g/mol) must be combusted (burned) to provide enough energy required to heat 1.00 qt of water described in (b)? (Answer: (a) 46.4 kJ/g; (b) QH20-317 kJ; (c) 6.83 g)

Explanation / Answer

a) From given value for energy it is clear that

1 mol(molar mass =44.06g/mol) of propane produce energy = 2046kJ. So,

44.06 g propane produce energy = 2046 kJ

1 g propane produce energy = 2046/44.06 kJ=46.4 kJ

b) Specific heat of water is 4.184J/g°C. This means

1 g water need 4.184 J energy to raise the temperature through 1°C.So,

946.4 g water need energy to raise the temperature through 1°C = 4.184*946.4 J =3960 J. Now,

We have to change the temperature from 20 to 100°C that is through 80°C.So

Energy needed to raise the temperature by 1°C= 3960 J

For 80°C energy needed = 3960*80J=316800J=317kJ

c)From reaction it is clear that

2046 kJ energy is produced from propane = 44.06 g

317 kJ energy is produced from propane = 44.06*317/2046g

= 6.83 g

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.