consider the reaction between mg s and hcl(aq) to produce aqueous magnesium chloride and hydrogen gas. How many liters of hydrogen gas at STP will be produced when 12.15 g of magnesium reacts with an excess of hydrochloric acid

11.2L is the volume of hydrogen gas produced.

Explanation:

The reaction of Mg with HCl is:

Mg + 2HCl → MgCl₂ + H₂

Based on the reaction the moles of Magnesium are equal to moles of hydrogen produced, that is:

Moles Mg -Molar mass: 24.305g/mol-:

12.15g Mg * (1mol / 24.305g) = 0.500 moles of Mg = Moles of H₂

Now, using PV = nRT, where:

P is pressure -1atm at STP-

V is volume -Our incognite-

n are moles -0.500 moles of H₂-

R is gas constant -0.082atmL/molK-

And T is temperature -273.15K-:

1atm*V = 0.500mol*0.082atmL/molK*273.15K

V = 11.2L is the volume of hydrogen gas produced.

the answer is 1 i believe