The reaction for the combustion of methane can be expressed as follows.
CH4 + 2O2 --> CO2 + 2H2O
We solve first for the amount of carbon dioxide in moles by dividing the given volume by 22.4L which is the volume of 1 mole of gas at STP.
moles of CO2 = (5.6 L) / (22.4 L/1 mole)
moles of CO2 = 0.25 moles
Then, we can see that every mole of carbon dioxide will need 1 mole of methane
moles methane = (0.25 moles CO2) x (1 moles O2/1 mole CO2)
= 0.25 moles CH4
Then, multiply this by the molar mass of methane which is 16 g/mole. Thus, the answer is 4 grams methane.
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To find the rate constant we can write a rate expression for the following reaction:
2A + B → C
A rate expression is written as some rate constant multiplied by the concentrations of the reactants, with each concentration raised to the power of the molar coefficient. [A] has a coefficient of 2, and [B] has a coefficient of 1. Therefore, we get the following rate expression:
rate = k[A]²[B]
We are given a table of values and we can enter the three variable to solve for k.
k = (rate)/([A]²[B])
k = (0.035)/((0.05)²(0.05))
k = 280
We can confirm if the value for k is correct by using another set of concentrations, along with the rate constant and solve for the rate.
rate = 280 [0.10]²[0.05]
rate = 280 (0.01)(0.05)
rate = 0.14
The value we solved for agrees with the rate provided in the table, therefore we know our value for the rate constant is correct which is k = 280.
