Answer:
[tex]82.6875gCO_2[/tex]
Explanation:
There is a chemistry category, but:
[tex]80g C * \frac{1mol}{12.01g C} = 6.6611 mol C[/tex]
[tex]60 gO * \frac{1mol}{16.00gO} = 3.75 molO[/tex]
find the ratio by dividing by smallest value:
[tex]\frac{6.6611}{3.75} = 1.776\\\frac{3.75}{3.75} = 1.0[/tex]
find and balance chemical formula:
[tex]2O + C > CO_2[/tex]
2 molO and 1 mol C is needed for a complete reaction, therefore the limiting reactant is the O
solve for CO2 using the mol of O
[tex]3.75molO * \frac{1molCO_2}{2molO} = 1.875molCO_2[/tex]
convert mol to g again
[tex]1.875molCO_2 * \frac{12.01+16.00*2}{1mol} = 82.6875 gCO_2[/tex]
