. How many grams of oxygen are required to burn 47.0g of C3H8?
how many moles in 47g of C3H8?
what is the reaction equation? That will tell you how many moles of O2 to use.
convert that back to grams.
C3H8 + 5O2 ==> 3CO2 + 4H2O is the equation you need.
To determine how many grams of oxygen are required to burn 47.0g of C3H8, you need to use the balanced chemical equation for the combustion of propane (C3H8). The balanced equation is:
C3H8 + 5O2 -> 3CO2 + 4H2O
From this equation, you can see that 3 moles of oxygen (O2) are required to burn 1 mole of propane (C3H8).
1 mole of propane (C3H8) has a molar mass of 44.1 g + 3(1.0 g) = 44.1 g + 3.0 g = 47.1 g.
To find the number of moles of propane in 47.0 g, you divide the mass of propane by its molar mass:
Number of moles of propane = 47.0 g / 47.1 g/mol = 1.0 mol
Since the ratio between propane and oxygen in the balanced equation is 1:5, you multiply the number of moles of propane by the ratio to find the moles of oxygen needed:
Number of moles of oxygen = 1.0 mol propane × (5 mol oxygen / 1 mol propane) = 5.0 mol
Finally, you can convert the moles of oxygen to grams by multiplying by the molar mass of oxygen, which is approximately 32.0 g/mol:
Grams of oxygen = 5.0 mol × 32.0 g/mol = 160 g
Therefore, 160 grams of oxygen are required to burn 47.0 grams of propane (C3H8).