Question

given the reaction C + H2O forms co+h2 process to calculate the volume of the gas produced at STP from 40.8 gram of carbon

Answer 1

first we need to balance the chemical equation you have given for CH4 + O2 --> CO2 + H2O

There are 4 hydrogens on the left so there should be 4 hydrogens on the right. That means we need a 2 in front of the water. As a result, we need to add a 2 in front of the O2 to balance out the oxygens. That gives the balanced equation as:

CH4 + 2O2 --> CO2 + 2H2O

Now let's convert the grams of methane and oxygen to moles and see which one is the limiting reagent.

8 g methane (1 mole/16 g) = 0.5 mole methane

15.9 g O2 (1 mole/32 g) = 0.5 mole O2

But from the equation we know that we need 2 moles of oxygen for every mole of methane so that makes oxygen the limiting reagent.  So 0.5 moles of O2 will form 0.25 moles of CO2 (2:1 ratio of O2:CO2).  So now convert moles of CO2 to grams:

0.25 moles CO2 (44 g/1 mole) = 11 grams CO2

So 11 grams of carbon dioxide are produced.