How many liters of hydrogen gas is produced at 298 K and 0.940 atm if 4.00 moles of hydrochloric acid react with an excess of magnesium metal?

The number of liters of hydrogen gas that is produced at 298 k and 0.940 atm if 4 moles of HCL reacted with an excess of magnesium metal is 52.05 L

calculation

step 1: calculate the number of moles of H2

write the equation for reaction Mg+2HCl → MgCl2 + H2 by use of mole ratio of HCl: H2 which is 2:1 the moles of H2 = 4.00 x1/2 = 2 moles

Step 2: use the ideal gas equation to calculate the volume of hydrogen

that is PV=nRT where

p (pressure) = 0.940 atm

V (volume) = ?

n (moles) = 2 moles

R (gas constant) = 0.0821 atm. l/mol. k

T (temperature) = 298 k

make V the subject of the formula

V is therefore=nRT/P

= [ (2 moles x 0.0821 atm. L/mol. k x 298 K) / 0.940 atm]

answer=52.05 L