C3H8+3O2--->3CO2+8H
Therefore for every 1:3 there are 3 Carbon dioxides that form. That means find the limiting reactant from the two reactants.
5.5g(1mole C3H8/44.03g of C3H8)=0.1249 moled of C3H8 and if for every one C3H8 we can form three CO2. We can assume 0.3747 miles of CO2 will be produced.
15g of O2(1 mole O2/32g of O2)=0.4685moles O2 and if for every three O2 we can produce three CO2 we may assume a 1:1 ratio.
This means C3H8 will be your limiting reactant. Therefore 0.3747 moles of CO2 will be produced.
0.3747 moles of CO2(48.01 g of CO2/1 mole of CO2)= 17.99 grams of CO2
Oxygen (6O2) and Glucose (C6H12O6)
<span>Reference: 6CO2 + 6H2O + light energy = C6H12O6 + 6O2.</span>
There is no reaction because of the common ion effect(which in this case is K)if there were a reaction the products would be the same as the reactants so no reaction happens
