If you have Hydrochloric acid, how many grams of aluminum must you use to produce 57.0 L of hydrogen at STP?
2Al(s)+6HCl(aq) --> 3H2(g)+2AlCl3(aq)
Just follow the steps in this example.
http://www.jiskha.com/science/chemistry/stoichiometry.html
I ended up getting 42.718333 not rounded to sig figs but does the excess of hydrochloric acid have anything to do with the answer and what about the STP?
I have something like 45. something and the problem says nothing about excess HCl; however, it won't work this way unless HCl IS IN excess.
You use the 57.0 L at STP to calculate moles H2. Remember that 1 mole of a gas at STP occupies 22.4 L; therefore, 57.0/22.4 = ?? moles H2.
Then ?? moles H2 x (2 moles Al/3 moles H2) = ?? moles H2 x (2/3) = moles Al and that times molar mass Al = about 45 g or so Al.
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To find out how many grams of aluminum are needed to produce 57.0 L of hydrogen gas at STP, we need to use stoichiometry.
First, let's convert the volume of hydrogen from liters to moles. We can use the ideal gas law to do this. At STP (Standard Temperature and Pressure), 1 mole of any gas occupies a volume of 22.4 liters.
So, 57.0 L of hydrogen gas is equal to 57.0 L / 22.4 L/mol = 2.54 moles of hydrogen.
Next, we need to use the balanced chemical equation and stoichiometry to determine the mole-to-mole ratio between aluminum and hydrogen gas. From the equation, we can see that 2 moles of aluminum react to produce 3 moles of hydrogen gas.
Therefore, the mole-to-mole ratio is 2 moles Al : 3 moles H2.
Now, we can set up a proportion to find the moles of aluminum required:
2 mol Al / 3 mol H2 = X mol Al / 2.54 mol H2
Cross-multiplying gives us:
3 * X = 2 * 2.54
3X = 5.08
X = 5.08 / 3
X ≈ 1.69 moles Al
Finally, we need to convert moles of aluminum to grams. The molar mass of aluminum (Al) is 26.98 g/mol.
So, the mass of aluminum required is:
Mass = Moles * Molar Mass
Mass = 1.69 mol * 26.98 g/mol
Mass ≈ 45.6 grams
Therefore, approximately 45.6 grams of aluminum are needed to produce 57.0 L of hydrogen gas at STP.