How many grams of iron are needed to produce 42.8 L of hydrogen gas according to the following reaction at 25oC and 1 atm?
iron (s) + hydrochloric acid (aq) iron(II) chloride (aq) + hydrogen (g)
Here is a worked example of a stoichiometry problem such as this. Just follow the steps. 42.8L/22.4 = moles H2 gas.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the number of grams of iron needed to produce 42.8 L of hydrogen gas, we need to use stoichiometry and the ideal gas law.
First, let's write the balanced chemical equation for the reaction:
Fe(s) + 2HCl(aq) -> FeCl2(aq) + H2(g)
From the balanced equation, we can see that one mole of iron reacts with 2 moles of hydrochloric acid to produce 1 mole of hydrogen gas.
Next, we need to convert the volume of hydrogen gas into moles. We can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (1 atm)
V = volume (42.8 L)
n = number of moles (to be determined)
R = ideal gas constant (0.08206 L·atm/(mol·K))
T = temperature (25°C or 298 K)
Rearranging the equation to solve for n:
n = PV / RT
n = (1 atm * 42.8 L) / (0.08206 L·atm/(mol·K) * 298 K)
n ≈ 1.776 moles
Since the stoichiometry of the reaction shows that 1 mole of iron reacts to produce 1 mole of hydrogen gas, we can conclude that we need 1.776 moles of iron.
The molar mass of iron is approximately 55.85 g/mol.
Finally, we can calculate the mass of iron needed:
Mass = moles * molar mass
Mass = 1.776 moles * 55.85 g/mol
Mass ≈ 99.18 grams
Therefore, approximately 99.18 grams of iron are needed to produce 42.8 L of hydrogen gas.