If 116g of butane burns in presence of 320g of oxygen, how much carbon dioxide will be produced? What is limiting reagent?
So This is how I worked it out:
116gC4H10*1/58.12(molar mass of C4H10)*8 moles CO2/2 moles C4H10*44/1(molar mass of CO2)=351.27g of CO2
Thank You!!
To determine the amount of carbon dioxide produced, we need to first determine the limiting reagent in the reaction. The limiting reagent is the reactant that is completely consumed and thus limits the amount of product formed.
To find the limiting reagent, we compare the number of moles of each reactant with their respective stoichiometric coefficients in the balanced chemical equation. The balanced chemical equation for the combustion of butane (C4H10) is:
2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O
Using the given information, we calculate the number of moles of each reactant:
Moles of butane (C4H10) = 116 g / molar mass of C4H10 = 116 g / 58.12 g/mol = 2 moles
Moles of oxygen (O2) = 320 g / molar mass of O2 = 320 g / 32 g/mol = 10 moles
Next, we compare the ratio of moles of butane to moles of oxygen based on the balanced equation:
Moles of butane / stoichiometric coefficient of butane = 2 moles / 2 = 1 mole
Moles of oxygen / stoichiometric coefficient of oxygen = 10 moles / 13 = 0.77 moles
Based on the comparison, we see that 0.77 moles of oxygen are required for each mole of butane. Since we have more oxygen than this ratio, the limiting reagent is the butane. Therefore, all the butane will be consumed in the reaction and determine the amount of carbon dioxide produced.
Now, we can calculate the amount of carbon dioxide produced using the stoichiometry from the balanced equation:
Moles of carbon dioxide (CO2) = Moles of butane * stoichiometric coefficient of CO2 / stoichiometric coefficient of butane
= 2 moles * 8 / 2 = 8 moles
Mass of carbon dioxide (CO2) = Moles of CO2 * molar mass of CO2
= 8 moles * 44 g/mol = 352 g
Therefore, when 116 g of butane burns in the presence of 320 g of oxygen, 352 g of carbon dioxide will be produced.