Annual <span> $105,200 but yealy $135,670
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the reaction is
2NO(g) + 2H2(g) <—> N2(g) + 2H2O (g)
Kc = [N2] [ H2O]^2 / [NO]^2 [ H2]^2
Given
moles of NO = 0.124 therefore [NO] = moles /volume = 0.124 /2 = 0.062
moles of H2 = 0.0240 , therefore [H2] = moles / volume = 0.0240 / 2 = 0.012
moles of N2 = 0.0380 , therefore [N2] = moles / volume = 0.0380 / 2 = 0.019
moles of H2O = 0.0276 , therefore [H2O] = moles / volume = 0.0276 / 2 = 0.0138
Kc = (0.019) ( 0.0138)^2 / (0.062)^2 ( 0.012)^2 = 6.54
Answer: <span>The actual yield depends on the reaction conditions, but the theoretical yield varies only with reactant amounts.
The actual yield has to be detemined from the real data.
The theoretical yield is calculated with the reactants amoounts and the theroretical molar ratios indicated by the chemical equation. Thus, the theoretical yield is the maximum possible yield, from the total reaction of the limiting reactant. The actual yield must be detemined under the particular contitions on which the reaction is performed.
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Answer:
The minimum mass of ethane that could be left over by the chemical reaction is zero.
Explanation:
Moles of ethane =
Moles of oxygen gas :
Since, 2 moles of ethane gas react with 7 moles of oxygen gas.
Then 0.7666 mol of ethane gas will react wit:
This means that ethane gas is in limited amount.hence ethane is a limiting reagent. Where as oxygen in present in excessive amount.
All the moles ethane gas will go under combustion reaction leaving zero moles behind.