Methanegasand oxygengas react to form carbon dioxide gas and watervapor. Suppose you have of and of in a reactor. Calculate the largest amount of that could be produced. Round your answer to the nearest .

Kindly check the explanation section.

Explanation:

NB: kindly take note that the question is not complete since we do not have any values for methane gas and water. I tried looking for the complete question but, I was unable to get it.

So, let's make use of the values of 7mols for Oxygen gas and 11 moles for the methane gas.

Therefore, the chemical equation for the reaction is given below:

CH4(g) + 2O2(g) > CO2(g) + 2H2O.

From the equation of reaction we have that 1 mole of Methane, CH4 reacts with 2 moles of Oxygen,O2 to give 2 moles of water, H2O and one mole of Carbondioxide,CO2. So, we have the ratio= 1 : 2 : 1 : 2.

The next thing for us to do is to find the limiting reagent which can be determine by calculating the number of moles In each reactants.

Number of moles of O2 = 7 moles

Number of moles of methane, CH4 = 11/1 = 11 moles.

Thus, the limiting reagent is Oxygen.

The largest amount of CO2 that could be produced is 7 moles of O2 × (1 mole of CO2/ 2 mol O2. = 3.5 moles of CO2.

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the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.