2 Fe (s) + 6 HCl (aq) -> 2 FeCL3 (s) + 3 H2 (g)
How many liters of H2 would be formed if 200 g of Fe completely reacted with HCl at 40 degrees Celsius and 1.5 atm?
http://www.jiskha.com/science/chemistry/stoichiometry.html
thanks!
To determine the number of liters of H2 gas that would be formed, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure in atm
V = volume in liters
n = number of moles
R = ideal gas constant (0.0821 L•atm/mol•K)
T = temperature in Kelvin
First, let's calculate the number of moles of Fe by using its molar mass:
Fe: 55.85 g/mol
Given:
Mass of Fe = 200 g
Number of moles of Fe = mass / molar mass = 200 g / 55.85 g/mol
Next, we need to find the limiting reactant between Fe and HCl. The balanced equation tells us that the ratio of the stoichiometric coefficients is 2:6 (Fe:HCl), which simplifies to 1:3. This means that every 1 mole of Fe reacts with 3 moles of HCl.
Number of moles of HCl = (number of moles of Fe) x (3 moles of HCl/1 mole of Fe)
Given:
Temperature (T) = 40 degrees Celsius = 313 K
Pressure (P) = 1.5 atm
Now, let's use the ideal gas law to calculate the volume of H2 gas produced:
PV = nRT
Rearranging the equation to solve for V:
V = (nRT) / P
Substituting the values into the equation:
V = [(number of moles of H2) x (0.0821 L•atm/mol•K) x (313 K)] / (1.5 atm)
Finally, plug in the calculated number of moles of H2 into the equation above to find the volume in liters.