What minimum amount of 1.55 M \rm HCl is necessary to produce 26.5 L of \rm H_2 at STP?
To determine the minimum amount of 1.55 M HCl needed to produce 26.5 L of H2 at STP, we can use the balanced chemical equation for the reaction between HCl and H2:
2 HCl(aq) -> H2(g) + Cl2(g)
The stoichiometry of the reaction tells us that 2 moles of HCl produce 1 mole of H2.
To find the number of moles of H2, we can use the ideal gas law at STP:
PV = nRT
At STP, the pressure (P) is 1 atm and the temperature (T) is 273.15 K.
Since we know the volume (V) of H2 is 26.5 L, we can rearrange the equation to solve for the number of moles (n) of H2:
n = PV / RT
Plugging in the values:
n = (1 atm) * (26.5 L) / (0.0821 L·atm/mol·K) * (273.15 K)
Simplifying the equation:
n = 1.0163 moles of H2
Since the stoichiometry of the reaction tells us that 2 moles of HCl react with 1 mole of H2, we need twice the number of moles of H2.
Therefore, the number of moles of HCl needed is:
2 * 1.0163 = 2.0326 moles of HCl
Now, we can calculate the minimum amount of 1.55 M HCl needed. The concentration (M) of a solution is defined as the number of moles of solute per liter of solution. In this case, the solute is HCl.
So, to calculate the minimum amount of HCl in liters, we can use the equation:
volume (L) = moles of solute / Molarity
volume (L) = 2.0326 moles / 1.55 mol/L
Simplifying the equation:
volume (L) = 1.3119 L
Therefore, the minimum amount of 1.55 M HCl necessary to produce 26.5 L of H2 at STP is approximately 1.3119 L.
mols H2 needed = 26.5L/22.4L = ?
Convert mols H2 to mols HCl. Then
MHCl = mols HCl/L HCl. You know M and mols solve for L HCl.