Therefore, the volume of hydrogen gas produced at STP when 23g of HCl reacts completely is approximately 7.056 L.
Explanation:
To calculate the volume of hydrogen gas produced at STP (Standard Temperature and Pressure) when 23g of HCl reacts completely, we need to use the concept of molar mass and molar volume.
1. Determine the molar mass of HCl:
- H (hydrogen) has a molar mass of 1g/mol
- Cl (chlorine) has a molar mass of 35.5g/mol
- The molar mass of HCl is therefore 1g/mol + 35.5g/mol = 36.5g/mol.
2. Calculate the number of moles of HCl:
- Given mass: 23g
- Number of moles = mass / molar mass = 23g / 36.5g/mol = 0.63 mol.
3. Apply the stoichiometry of the balanced chemical equation to determine the number of moles of hydrogen gas produced. The balanced chemical equation for the reaction between HCl and hydrogen gas (H2) is:
2HCl + 2H2O -> 2H3O+ + Cl2
From the balanced equation, we see that 2 moles of HCl produce 1 mole of H2.
Therefore, 0.63 mol of HCl will produce 0.63/2 = 0.315 mol of H2.
4. Use the molar volume of a gas at STP to calculate the volume of hydrogen gas produced:
- The molar volume of a gas at STP is approximately 22.4 L/mol.
- Volume of H2 gas = number of moles x molar volume = 0.315 mol x 22.4 L/mol = 7.056 L.
Therefore, the volume of hydrogen gas produced at STP when 23g of HCl reacts completely is approximately 7.056 L.