Methane reacts with oxygen to produce carbon dioxide, water and heat. If the percent yield of the reaction is 90.0%, how many grams of methane must be burned to produce 150.0 kJ of energy?
___ CH4 (g)+ ___ O2 (g) → ___ CO2 (g) + ___ H2O(g) + 802 kJ
CH4(g)+ 2O2(g) → CO2(g) + 2H2O(g) + 802 kJ
You obtain 802 kJ for every 16 g CH4 so
802 kJ/mol x (g/16) = 150 kJ.
Solve for g
To find out how many grams of methane must be burned, we need to use the given information about the reaction and the percent yield.
First, let's determine the balanced equation for the reaction:
CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O(g) + 802 kJ
From the balanced equation, we can see that 1 mole of methane (CH4) reacts to produce 1 mole of carbon dioxide (CO2), 2 moles of water (H2O), and releases 802 kJ of energy.
Next, we can calculate the energy produced by burning 1 mole (16.04 g) of methane:
802 kJ/1 mole CH4 = 802 kJ/16.04 g CH4 = 50 kJ/g CH4
Now, let's calculate the amount of energy produced by burning the desired amount of methane (150.0 kJ):
150.0 kJ × (1 mole CH4/50 kJ) = 3 moles CH4
Since we are given that the percent yield of the reaction is 90.0%, we need to account for the efficiency of the reaction. The percent yield tells us that only 90.0% of the expected amount of product is actually produced.
So, we can calculate the actual amount of methane needed to produce 150.0 kJ of energy with the given percent yield:
3 moles CH4 × (100%/90%) = 3.33 moles CH4
Finally, we can convert moles of methane to grams using the molar mass of methane:
3.33 moles CH4 × 16.04 g/mol = 53.352 g CH4
Therefore, approximately 53.4 grams of methane must be burned to produce 150.0 kJ of energy with a percent yield of 90.0%.