#C_2H_4 + 3O_2 - 2CO_2 +2H_2O#. If you start with 45 grams of ethylene (#C_2H_4#), how many grams of carbon dioxide will be produced?

Question

#C_2H_4 + 3O_2 -> 2CO_2 +2H_2O#.  If you start with 45 grams of ethylene (#C_2H_4#), how many grams of carbon dioxide will be produced?

Wyatt Atkins · Accepted Answer

141.416g

#"No. of mol of ethylene"=45/(2*12+4*1)~~1.607mol# #"No. of mol of "CO_2" produced"=2*1.607=3.214mol# #"Mass of "CO_2"=3.214*(12+2*16)=141.416g#

Cooper Adams · Answer

undefined To find the mass of carbon dioxide produced, first, calculate the molar mass of ethylene (C₂H₄) and carbon dioxide (CO₂).

Molar mass of C₂H₄:
C: 12.01 g/mol (2 atoms)
H: 1.01 g/mol (4 atoms)

Total molar mass = (2 * 12.01 g/mol) + (4 * 1.01 g/mol) = 28.05 g/mol

Then, use stoichiometry to determine the ratio of moles of ethylene to moles of carbon dioxide:

1 mol of C₂H₄ produces 2 mol of CO₂.

Now, calculate the number of moles of ethylene:

Given mass of ethylene = 45 g
Molar mass of ethylene = 28.05 g/mol

Number of moles of ethylene = 45 g / 28.05 g/mol

Now, use the mole ratio to find the number of moles of CO₂ produced:

Number of moles of CO₂ = Number of moles of ethylene * 2

Finally, find the mass of carbon dioxide produced using the molar mass of CO₂:

Mass of CO₂ = Number of mo

#C_2H_4 + 3O_2 -&gt; 2CO_2 +2H_2O#. If you start with 45 grams of ethylene (#C_2H_4#), how many grams of carbon dioxide will be produced?