Aluminum carbide, Al2C3 (s), reacts with water to yield aluminum hydroxide, Al(OH)3 (aq), and methane, CH4 (g). Write the equation for the reaction. How many grams of Al4C3 (s) would be needed to prepare 1.000 L of CH4 (g) measured at STP?
write, balance the equation.
one liter of the gas is 1/22.4 moles.
Now look at the mole ratio: how many moles of Al2C3 does it take? Change that to grams.
To write the equation for the reaction, we need to balance it first.
The reaction is as follows:
Al2C3 (s) + 12 H2O (l) → 2 Al(OH)3 (aq) + 3 CH4 (g)
Now, let's move on to calculating the number of grams of Al4C3 (s) needed to prepare 1.000 L of CH4 (g) at STP.
STP stands for Standard Temperature and Pressure, which is defined as 0 degrees Celsius and 1 atmosphere of pressure.
Step 1: Convert 1.000 L of CH4 (g) to moles
To convert from liters to moles, we need to know the molar volume of a gas at STP, which is approximately 22.4 L/mol. Therefore, 1.000 L of CH4 (g) at STP is equal to 1.000/22.4 = 0.0446 moles of CH4.
Step 2: Use the stoichiometry of the balanced equation to find the moles of Al4C3 (s)
From the balanced equation, we can see that 3 moles of CH4 are produced for every 1 mole of Al4C3. Therefore, we can set up the following equation:
0.0446 moles CH4 × (1 mole Al4C3 / 3 moles CH4) = 0.0149 moles Al4C3
Step 3: Calculate the mass of Al4C3 (s)
The molar mass of Al4C3 is approximately 144 g/mol. Therefore, the mass of Al4C3 needed is:
0.0149 moles Al4C3 × 144 g/mol = 2.14 grams of Al4C3 (rounded to two decimal places)
So, approximately 2.14 grams of Al4C3(s) would be needed to prepare 1.000 L of CH4(g) at STP.