In the actual experiment 146.2 mL of O2(g) was collected. This is because the O2(g) was collected over water and the atmospheric pressure includes both the pressure of oxygen and water vapor. Use Dalton’s Law of Partial Pressures and Boyle’s Law to correct for the water vapor pressure and determine the volume of dry oxygen.

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2

un recipient conține 6,4 moli de gaz. gazul hidrogen reprezintă 25% din totalul kilometrilor din container. în cazul în care presiunea totală este de 1.24 atm ceea ce este