The number of liters of H2 that would be formed at 600 mmhg and 13 c if 26.5 g zinc was allowed to react is calculated as follows
find the moles of Zn reacted
that is 26.5 g / 56.39 g/mol = 0.4699 moles
write the reaction Zn with an acid example HCl to produce H2
that is Zn + 2 HCl = ZnCl2 + H2
by use of mole ratio between Zn to H2 the moles of H2 = 0.4699 moles
By use of Ideal gas equation Volume of H2 =nRT/P
n= 0.499 moles
R= 62.364 L.mmhg/mol.K
T=13 +273=286 k
P= 600 mmhg
= (0.4699 mol x 62.364 L.mmhg/mol.k x286 k)/600 mm hg = 13.97 liters of H2
