What is the minimum volume of 0.1M barium hydroxide required to absorb the carbon dioxide produced by burning 32g of methane?
What is your answer?
I got 10-Liters.
To find the minimum volume of 0.1M barium hydroxide required to absorb the carbon dioxide produced by burning 32g of methane, we need to follow these steps:
Step 1: Write the balanced chemical equation for the combustion of methane.
CH4 + 2O2 -> CO2 + 2H2O
Step 2: Calculate the molar mass of methane (CH4).
Molar mass of carbon (C) = 12.01 g/mol
Molar mass of hydrogen (H) = 1.01 g/mol
Molar mass of methane (CH4) = (4 x molar mass of hydrogen) + molar mass of carbon
= (4 x 1.01) + 12.01
= 16.05 g/mol
Step 3: Convert the mass of methane (32g) to moles.
Number of moles = mass / molar mass
= 32 / 16.05
= 1.992 moles (approximately)
Step 4: Determine the stoichiometry of the balanced equation.
From the balanced equation, we can see that 1 mole of methane produces 1 mole of carbon dioxide. Therefore, 1.992 moles of methane will produce 1.992 moles of carbon dioxide.
Step 5: Calculate the volume of 0.1M barium hydroxide required.
Since we know the concentration of the barium hydroxide solution and the balanced equation, we can use the formula:
Volume of solution (in liters) = Number of moles / Concentration
Volume of barium hydroxide = 1.992 moles / 0.1 mol/L
= 19.92 liters
Therefore, the minimum volume of 0.1M barium hydroxide required to absorb the carbon dioxide produced by burning 32g of methane is approximately 19.92 liters.