2) Given the following equation:
Al2(SO3)3 + 6 NaOH > 3 Na2SO3 + 2 Al(OH)3

a) If 10.0 g of Al2(SO3)3 is reacted with 10.0 g of NaOH, determine the limiting reagent
b) Determine the number of moles of Al(OH)3 produced
c) Determine the number of grams of Na2SO3 produced
d) Determine the number of grams of excess reagent left over in the reaction

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> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.

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