The balanced equation below represents the reaction of glucose, C 6 M 12 O 4 oxygen at 298 K and 101.3 kPa C 6 H 12 O 6 (s)+6O 2 (g) 6CO 2 (g)+6H 2 O(l) Determine the mass of CO 2 produced when 9,0 grams of glucose completely reacts with 9.6 grams of oxygen to produce 5.4 grams of water.

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