Answer:
12.44 g
Explanation:
2C4H10 + 13O2 = 8CO2 + 10H2O
n(C4H10) = m(C4H10)/M(C4H10) = 4.1 / 58g/mol = 0.0707 mol (excess).
n(O2) = m(O2)/M(O2) = 25.9 / 32g/mol = 0.809 mol (deficiency).
Since the ratio of O2 to octane is 13 : 2 we can divide 0.0707 by 2 to get 0.03535 and divide 0.809 by 13 to get 0.062.
mass of CO2 produced =
M = [0.0707 moles C4H10 x 8 moles CO2] / 2 moles C4H10 x 44 g CO2/mol
M = 0.5656/2 * 44
M = 0.2828 * 44
M = 12.44 of CO2
Answer: The ratio of carbon dioxide molecules to oxygen molecules is 20 :31
Explanation:
Combustion is a chemical reaction in which hydrocarbons are burnt in the presence of oxygen to give carbon dioxide and water.
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
The balanced combustion reaction is,:
![2C_{10}H_{22}(l)+31O_2(g)\rightarrow 20CO_2(g)+22H_2O(g)](https://tex.z-dn.net/?f=2C_%7B10%7DH_%7B22%7D%28l%29%2B31O_2%28g%29%5Crightarrow%2020CO_2%28g%29%2B22H_2O%28g%29)
Thus the ratio of carbon dioxide molecules to oxygen molecules is 20 : 31
You can use the equation ΔS(surr)=q(surr)/T or ΔS(surr)=-q(rxn)/T.
the two equations are equal since we know that the energy the system (reactoin) puts out just goes into the surroundings.
(In other words q(surr)=-q(rxn))
Using the equation, <span>ΔS(surr)=-(-283kJ/298K)=0.9497kJ/K or 949.7J/K
This answer makes sense since the reaction is exothermic which means it released energy into the system which usually causes the entropy to increase.
I hope that helps.</span>