2 Al + 2 NaOH + 2H2O to 2 NaAlO2 + 3H2
if 126.4 g of NaOH is 97.7 g of Al are made of to react
a. How
many moles of NaAlO2 is produced if 126,4 g of NaOH reacts completely?
b. How many grams of NaAlO2 is produced using the same amount of NaOH?
c. Which reactant is the limiting reagent?
d How many grams of the excess reagent is left?
a)Moles of NaAlO2=4.97moles
b) Mass of NaAlO2 =116.8g
c)Limiting reagent=Al
d) Excess reagent =24.8g
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