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
Omg hard question but it’s also amazing imma think about that question
Answer : The mass of oxygen will be, 36.8 grams
Solution : Given,
Moles of water = 2.30 moles
Molar mass of 
= 32 g/mole
First we have to calculate the moles of oxygen.
The balanced chemical reaction will be,
From the balanced reaction we conclude that
2 moles of water decomposes to give 1 mole of
2.30 moles of water decomposes to give 
moles of
Now we have to calculate the mass of oxygen.
Therefore, the mass of oxygen will be, 36.8 grams
