The chemical reaction for this is:
2 C2H6 + 7 O2 => 4 CO2 + 6 H2O
Solving for CO2 with each reactant will give:
21.0 g C2H6 x (1 mol C2H6/30.08 g C2H6) x (6 mol H2O/2
mol C2H6) x (18 g H2O/1 mol H2O) = 37.70 g H2O
110 g O2 x (1 mol O2/32.00 g O2) x (6 mol CO2/7 mol O2) x
(18 g H2O/1 mol H2O) = 53.04 g H2O
Since the amount of H2O in C2H6 is lower therefore C2H6
is the limiting reactant and the maximum amount of water is only 38 g H2O (2 significant digits)
ANswer:
38 g water
Answer:
![K=K_1*K_2\\\\K=\frac{[H_2]^3[CO_2][H_2]}{[CH_4][H_2O][H_2O]}](https://tex.z-dn.net/?f=K%3DK_1%2AK_2%5C%5C%5C%5CK%3D%5Cfrac%7B%5BH_2%5D%5E3%5BCO_2%5D%5BH_2%5D%7D%7B%5BCH_4%5D%5BH_2O%5D%5BH_2O%5D%7D)
Explanation:
Hello there!
In this case, for the given chemical reaction, it turns out firstly necessary to write the equilibrium expression for both reactions 1 and 2:
![K_1=\frac{[CO][H_2]^3}{[CH_4][H_2O]} \\\\K_2=\frac{[CO_2][H_2]}{[CO][H_2O]}](https://tex.z-dn.net/?f=K_1%3D%5Cfrac%7B%5BCO%5D%5BH_2%5D%5E3%7D%7B%5BCH_4%5D%5BH_2O%5D%7D%20%5C%5C%5C%5CK_2%3D%5Cfrac%7B%5BCO_2%5D%5BH_2%5D%7D%7B%5BCO%5D%5BH_2O%5D%7D)
Now, when we combine them to get the overall expression, we infer these two are multiplied to get:
![K=K_1*K_2\\\\K=\frac{[CO][H_2]^3}{[CH_4][H_2O]} *\frac{[CO_2][H_2]}{[CO][H_2O]}\\\\K=\frac{[H_2]^3[CO_2][H_2]}{[CH_4][H_2O][H_2O]}](https://tex.z-dn.net/?f=K%3DK_1%2AK_2%5C%5C%5C%5CK%3D%5Cfrac%7B%5BCO%5D%5BH_2%5D%5E3%7D%7B%5BCH_4%5D%5BH_2O%5D%7D%20%2A%5Cfrac%7B%5BCO_2%5D%5BH_2%5D%7D%7B%5BCO%5D%5BH_2O%5D%7D%5C%5C%5C%5CK%3D%5Cfrac%7B%5BH_2%5D%5E3%5BCO_2%5D%5BH_2%5D%7D%7B%5BCH_4%5D%5BH_2O%5D%5BH_2O%5D%7D)
Regards!
it would be heating of 32g of s and 56g of Fe to produce 88g of FeS and unused reactants
Ethylene is the starting material for the preparation of a number of two-carbon compounds including ethanol (industrial alcohol), ethylene oxide (converted to ethylene glycol for antifreeze and polyester fibres and films), acetaldehyde (converted to acetic acid), and vinyl chloride (converted to polyvinyl chloride).
