How many grams of CO2 can be produced from the combustion of 10 moles of O2? please anwser soon original question; If you have 10 moles of oxygen (O2), how many grams of CO2 are produced during this combustion? Solve and record your answer
using this equation
C3H8+502->3CO2+4H2O

The first thing you need to do is to write the correctly balanced equation for the combustion of C2H2.

2C2H2 + 5O2 ==> 4CO2 + 2H2O  ...   balanced equation

Next, we find which reactant, if any, is limiting.  There are several ways to do this, but one easy way is to find moles of each reactant, and divide that value by the coefficient in the balanced equation and see which is less.

For C2H2, we have...

1.26 g x 1 mole/26.04 g = 0.0484 moles C2H2 (÷2 = 0.0242)

For O2, we have...

3.39 g O2 x 1 mol/32 g = 0.1059 moles O2 (÷5 = 0.0211) <- This is less than 0.0242 so it is limiting

Next, we use the moles of the limiting reactant (0.1059, not 0.0211) to find moles and then grams of CO2:

0.1059 moles O2 x 4 moles CO2/5 mole O2 x 44 g CO2/mole CO2 = 3.73 g CO2

Explanation:

in scientific notation, the value of the  speed of light in a vacuum

  will be written as 2.995 * 10^8. when this value is represented in the form of

  a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.