How many grams of methane gas (CH4) need to be combusted to produce 18.2 L water vapor at 1.2 atm and 275 K? Show all of the work used to solve this problem.

CH4 (g) + 2 O2 (g) yields CO2 (g) + 2 H2O (g)

Question

How many grams of methane gas (CH4) need to be combusted to produce 18.2 L water vapor at 1.2 atm and 275 K? Show all of the work used to solve this problem.

CH4 (g) + 2 O2 (g) yields CO2 (g) + 2 H2O (g)

anonymous · Answer

How many liters of 13.5 molar HCl stock solution are needed to make 10.5 liters of a 1.8 molar HCl solution? Show the work used to solve this problem.

zepp · Answer

2 questions in one? :|

anonymous · Answer

Sorry, I can separate them so you can get 2 medals?

zepp · Answer

No no, it's okay, lemme see.

zepp · Answer

And by 275 K you mean 275 kilojoules right?

anonymous · Answer

I actually have no idea....

anonymous · Answer

I'll just skip it

anonymous · Answer

lets just do the second one

zepp · Answer

lol ok

zepp · Answer

So from what I understand, you need to make a dilution in the second problem.

anonymous · Answer

Okay..

zepp · Answer

Um, I forgot how to dilution stuffs, I'm re-reading my Chem book, will be back with the solution shortly :$

zepp · Answer

Ok

zepp · Answer

Let's identify our variables.

We have a stock solution (HCL) concentrated at 13.5M/L

We need a 10.5 liters of a 1.8 molar HCl solution

zepp · Answer

We'll be using the formula $C_1V_1=C_2V_2$ to find our solution.

zepp · Answer

So, our $C_1$ would be $13.5mol/L$

$C_2=1.8mol/L$
$V_2=10.5L$

We need to find how many liters of the stock solution of HCl we need

$C_1V_1=C_2V_2\13.5M/L*V_1=1.8M/L * 10.5L\13.5M/L*V_1=18.9M\V_1=18.9M \div 13.5M/L=1.4L$