How many liters of h2 would be formed at 600 mm hg and 13 ∘c if 26.5 g of zinc was allowed to react?

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