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 reactionMg+2HCl → MgCl2  + H2by 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

High-energy and unstable.

Answer trigonal pyramidal;