In the equation #2C_2H_6 + 7O_2 - 4CO_2 + 6H_2O#, if 15 g of #C_2H_6# react with 45 g of #O_2#, how many grams of water will be produced?

Question

In the equation #2C_2H_6 + 7O_2 -> 4CO_2 + 6H_2O#, if 15 g of #C_2H_6# react with 45 g of #O_2#, how many grams of water will be produced?

Ellie Andrews · Accepted Answer

#"Moles of ethane"# #=# #(15*g)/(30.07*g*mol^-1)=0.499*mol#.

I gets #27*g# water product............

#"Moles of dioxygen"# #=# #(45*g)/(32.00*g*mol^-1)=1.41*mol#. Clearly, there is insufficient dioxygen for complete combustion; i.e. complete combustion requires #1.75*mol# #O_2#. We ASSUME incomplete combustion, i.e. #"COMPLETE COMBUSTION GIVES...."# #C_2H_6(g) + 7/2O_2(g) rarr 2CO_2(g) + 3H_2O(l)# #"INCOMPLETE COMBUSTION GIVES...."# #C_2H_6(g)+3O_2(g)rarrCO_2(g) + CO(g)+3H_2O(l)# #"OR.........."# #C_2H_6(g)+2O_2(g)rarrC(s)+ CO(g)+3H_2O(l)# (What would we obtain? This would be the experiment's focus.) Nevertheless, whether ethane is burned completely or incompletely, it always results in the following in both cases:... #"THREE EQUIV of WATER PER EQUIV ETHANE."# And thus if there are #0.5*mol# ethane reactant, there will be #1.5*mol# water product.........a mass of #27*g#. Excellent query, which I'm taking for granted for my A2 class.

Natalie Anton · Answer

undefined Once the limiting reactant has been identified, use stoichiometry to calculate the moles of water produced, and finally convert moles of water to grams using its molar mass. To find the grams of water produced, first calculate the moles of each reactant using their molar masses. Next, identify which reactant is limiting by comparing the moles of each reactant to the stoichiometric coefficients in the balanced equation.

In the equation #2C_2H_6 + 7O_2 -&gt; 4CO_2 + 6H_2O#, if 15 g of #C_2H_6# react with 45 g of #O_2#, how many grams of water will be produced?

In the equation #2C_2H_6 + 7O_2 -> 4CO_2 + 6H_2O#, if 15 g of #C_2H_6# react with 45 g of #O_2#, how many grams of water will be produced?