If 26.5 mL of 0.222 M HCl reacts with excess Mg, how many mL of hydrogen gas are produced at STP?

duplicate post.

26.5 *10^-3 L * 0.222 mol / L H+ = 8.1*10^-3 mol H+ or 4.05*10^-3 mol H2

4.05*10^-3 mol H2 * 22.4 Liters/mol at STP = 90.7 *10^-3 liters
= 90.7 mL

calc error

26.5 *10^-3 L * 0.222 mol / L H+ = 5.88*10^-3 mol H+ or 2.94*10^-3 mol H2
2.94*10^-3 mol H2 * 22.4 Liters/mol at STP = 65.9 *10^-3 liters
= 65.9 mL

Nope. I suspect you punched in the wrong numbers since 26.5E-3 x 0.222 = 5.88*10^-3 and half that is 2.94*10^-3 mols H2 gas @ STP

To determine the number of milliliters (mL) of hydrogen gas produced at STP, we need to use stoichiometry. Stoichiometry is a way to calculate the quantities of reactants and products in a chemical reaction.

First, we need to find the moles of HCl used in the reaction. To do this, we can use the equation:

moles of HCl = concentration of HCl (M) x volume of HCl (L)

Converting the volume of HCl from mL to L by dividing by 1000, we have:

moles of HCl = 0.222 M x (26.5 mL / 1000 mL/L) = 0.005883 moles of HCl

Since the stoichiometric ratio between HCl and H2 is 2:1 (for every 2 moles of HCl, 1 mole of H2 is produced), we can determine the number of moles of H2 produced:

moles of H2 = 0.005883 moles of HCl / 2 = 0.002941 moles of H2

Now, we can use the ideal gas law to calculate the volume of H2 at STP. At STP (Standard Temperature and Pressure), the volume of 1 mole of any gas is 22.4 L. So, using the ideal gas law:

volume of H2 = moles of H2 x molar volume at STP
volume of H2 = 0.002941 moles of H2 x 22.4 L/mol = 0.06594 L

To convert the volume of H2 to milliliters (mL), we multiply by 1000:

volume of H2 = 0.06594 L x 1000 mL/L = 65.94 mL

Therefore, approximately 65.94 mL of hydrogen gas are produced at STP when 26.5 mL of 0.222 M HCl reacts with excess Mg.