When carbon is burned in air carbon iv oxide gas is formed.
C (s) + O2 (g) = CO2(g) ΔH = - 393.5 kj/mol
The enthalpy change of the reaction is -393.5 j/mol which means that when one mole of carbon is completely burnt in air then 393.5 j of energy is evolved.
Thus, 1 mole = -393.5 j , then for 480 kj
= 480 × 1/393.5
= 1.2198 moles
1 mole of carbon iv oxide is equal to 44 g
thus, 1.2198 moles will be 1.2198 × 44 = 53.6712 g of CO2
Answer:
13.44dm^3
Explanation:
To calculate this we first need to know the number of mole produced. We will first need to balance the equation to know the theoretical mole ratio.
C2H6 (g) + 3.5O2 (g) → 2CO2 (g) + 3H2O (g)
From the balanced equation, we can deduce that one mole of ethane yielded 2 moles of carbon iv oxide. We use this information to calculate the actual number of moles yielded.
24g were reacted. Now to know the number of moles reacted, we simply divide the mass by the molar mass. The molar mass of ethane is 2(12) +6(1)= 40g/mol
The number of moles is thus 24/40 = 0.6 moles
Like we said earlier, one mole yielded 2 moles of carbon iv oxide, hence, 0.6 moles will yield 0.6 * 2 = 0.12 moles of carbon iv oxide.
Now, at stp, one mole of a gas occupies a volume of 22.4dm^3 thus, 0.6 mole will occupy 0.6 * 22.4 = 13.44dm^3
Answer: 7.2418 x 10^-19 joules
Explanation:
1 eV equals 1.602 x 10^-19 joules
Then 4.52eV will be multiplied by the above value to give 7.2418 x 10^-19 joules which is the energy required to dissociate the hydrogen molecule
