If 18.80 grams of hydrochloric acid are reacted with a large amount of aluminum metal, what volume of hydrogen gas would be produced at STP? (Hint: start by writing a balanced equation for the reaction.)
3HCl + Al -> 3H + AlCl3
18.80g
moles= grams/mw
18.80/109.5
0.1716894977169
I do not know if I have correctly balanced the equation, and I think I need to use PV=NRT somewhere, but I'm not sure how to get there.
2Al + 6HCl ==> 3H2 + 2AlCl3
One error you continue to make is to determine mole from the entire equation. moles = grams/molar mass. and NOT grams/n*molar mass.
Here is a worked example of a stoichiometry problem. Just follow the steps. That will show you how to do moles, too.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the volume of hydrogen gas produced at STP, we first need to determine the number of moles of hydrogen gas produced.
From the balanced equation:
3HCl + Al -> 3H + AlCl3
We can see that 3 moles of hydrogen gas are produced for every 3 moles of HCl reacted. Therefore, the number of moles of hydrogen gas produced is equal to the number of moles of HCl reacted.
You have correctly calculated the number of moles of HCl reacted as 0.1717 moles.
Now, using the ideal gas law equation PV = nRT, we can calculate the volume of hydrogen gas produced.
P is the pressure, which we assume to be standard pressure at STP, equal to 1 atm.
V is the volume, which is what we need to find.
n is the number of moles of hydrogen gas, which is 0.1717 moles.
R is the ideal gas constant, which is 0.0821 L·atm/(mol·K).
T is the temperature, which is also assumed to be at STP, equal to 273.15 K.
Rearranging the ideal gas law equation to solve for V, we have:
V = (nRT) / P.
Substituting the known values into the equation, we get:
V = (0.1717 moles * 0.0821 L·atm/(mol·K) * 273.15 K) / 1 atm.
Calculating this expression, we find:
V = 3.736 L.
Therefore, approximately 3.736 liters of hydrogen gas would be produced at STP when 18.80 grams of hydrochloric acid react with a large amount of aluminum metal.