Methane burns in the presence of oxygen to form carbon dioxide and water. 2nd attempt
A sealed chamber contains 7.50 g CH4 and 13.5g 02.
The mixture is ignited producing CO2 and H20. How many grams of CO2 are produced?

= 9.28 g CO₂

Explanation:

First write a balanced equation:

CH₄ + 2O₂ -> 2H₂O + CO₂

Convert the information to moles

7.50g CH₄ = 0.46875 mol CH₄

13.5g O₂ = 0.421875 mol O₂

Theoretical molar ratio CH₄:O₂ -> 1:2

Actual ratio is  0.46875 : 0.421875 ≈ 1:1

If all CH₄ is used up, there would need to be more O₂

So O₂ is the limiting reactant and we use this in our equation

Use molar ratio to find moles of CO₂

0.421875 mol O₂ * 1 mol CO₂/2 mol O₂=0.2109375 mol CO₂

Then convert to grams

0.2109375 mol CO₂ = 9.28114 g CO₂

round to 3 sig figs

= 9.28 g CO₂

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml