Please help me solve this
 

I am having alot of trouble with this equation and hope someone can help


The combustion of methane is represented by the equation

CH4 + 2 O2(g)--> CO2(g) + 2 H2O(l) ΔH = -890.3 kJ


What is q for the combustion of 105 L of CH4(g), measured at 23 °C and 746 mmHg?

First you have to figure out how many mols you got in the 105 L in this environment.

p*V = n*R*T

p = the pressure in Pa, so we have to convert. 746 mmHg = 99458.48 Pa
V = the volume this is 105 L which is 105 dm³ but we must have it in m³ so it's 0.105 m³.
n = the thing we are looking for.
R = It's a constant. 8.3145 m³·Pa/(mol·K)
T = the temperature in Kelvin! So if we have 23 degrees Celsius, we have 296.15 degrees in Kelvin.

Great! So we have:
(99458.48*0.105)/(8.3145*296.15) = 4.241 mol of CH4.

The equation says, when burnin 1 mole of CH4 890.3 kJ of heat are emitted into the environment. That is why it is an exotermic reaction. So if we have 4.241 mol of CH4, this will emit 4.241*890.3 of heat.
Q= 3775.76 kJ heat.