Determine how much CH4, O2, CO2, and H2O remain after the reaction is complete?

Question

1)Suppose 8.00g of CH4 is allowed to burn in the presence of 6.00g of oxygen. How much (in grams) CH4, O2, CO2, and H2O remain after the reaction is complete?

I have CH4 - 16g/mol , O2 - 32g/mol , CO2 - 42g/mol , H2O 20g/mol

I know that this question has been asked earlier but they explained it but I could not understand the steps. So if someone could show me the steps on how to solve this I would love it. I have a test tomorrow and a question is going to be like this on it.

And this question I only ask for steps on how to do it. I want to find the answer and learn. I can figure out A but need help on B


2)Consider the combustion of methane, CH4:
CH4(g)+2O2(g) -->CO2(g)+2H2O(g)
Suppose 2.8 moles of methane are allowed to react with 3 moles of oxygen.
A)what is the limiting reagent
B)How many moles of CO2 can be made from this mixture? How many grams of CO2?

Answer 1

CH4(g) + 2O2(g) --> CO2(g) + 2H2O(g)
1 mole CH4 needs 2 moles of O2 for complete combustion, and forms 1 mole of CO2 and 2 moles H20
Since the available O2 is limiting this reaction, to find the CO2 produced, set up the ratio:
3/2 = x/1
(the denominators are the coefficients of the chemicals in the equation)
x = 1.5 moles CO2
mass of 1.5 moles CO2 = 1.5*12*(16)*2 = 578 g

This, of course assumes complete combustion and no formation of CO, which can happen if the excess CH4 is not controlled