What minimum volume of 3.0 M H2SO4 is required to produced 13 L of H2 at STP?
Reaction formula:
2Al(s) + 3H2SO4(aq) --> Al2(SO4)3(aq) + 3H2 (g)
Here is a worked example of a stoichiometry problem and it will work all of your stoichiometry problems. Remember 13LH2/molar mass H2 = moles H2 and moles H2SO4 = M x L.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To find the minimum volume of 3.0 M H2SO4 required to produce 13 L of H2 at STP, we can use the balanced chemical equation and stoichiometry.
The balanced chemical equation shows that 3 moles of H2 are produced for every 3 moles of H2SO4 reacted. Therefore, the molar ratio between H2 and H2SO4 is 1:1.
The given volume of H2 is 13 L. Since the reaction is at STP (standard temperature and pressure), we can use the ideal gas law to determine the number of moles of H2 produced:
PV = nRT
Where:
P = pressure (at STP, P = 1 atm)
V = volume of gas (13 L)
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (at STP, T = 273 K)
Rearranging the equation to solve for n, we have:
n = PV / RT
Substituting the values into the equation:
n = (1 atm) * (13 L) / (0.0821 L·atm/mol·K * 273 K)
n ≈ 0.565 moles
Since the molar ratio between H2 and H2SO4 is 1:1, we need 0.565 moles of H2SO4 to produce 0.565 moles of H2.
The concentration of the H2SO4 solution is given as 3.0 M, which means there are 3.0 moles of H2SO4 in 1 liter of solution.
To calculate the minimum volume of 3.0 M H2SO4 required, we can use the equation:
Volume (in liters) = moles / concentration
Volume = 0.565 moles / 3.0 M
Volume ≈ 0.188 L
Therefore, the minimum volume of 3.0 M H2SO4 required to produce 13 L of H2 at STP is approximately 0.188 liters.