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
reactant and explain write the calculations
Answers
Al2(SO3)3 + 6NaOH = 3Na2SO3 + 2Al(OH)3
(a) moles Al2(SO3)3 = 10.0 g x 1 mol/294 g = 0.0340 moles Al2(SO3)3
moles NaOH = 10.0 g x 1 mol/40 g = 0.25 moles NaOH
Divide each by the coefficient in the balanced equation:
0.0340 Al2(SO3)3 /1 = 0.0340 moles
0.25 NaOH/6 = 0.0417 moles
Al2(SO3)3 is limiting as it is in the least supply
(b) 0.0340 moles Al2(SO3)3 x 2 moles Al(OH)3/mole Al2(SO3)3 = 0.0680 moles Al(OH)3 produced
(c) 0.0340 moles Al2(SO3)3 x 3 moles Na2SO3/mole Al2(SO3)3 = 0.102 moles Na2SO3
0.102 moles Na2SO3 x 126 g/mol = 12.9 g Na2SO3 produced
(d) Excess reagent is NaOH
moles NaOH used = 0.0340 moles Al2(SO3)3 x 6 moles NaOH/mole Na2(SO3)3 = 0.204 moles used
moles NaOH remaining = 0.25 - 0.204 = 0.046 moles NaOH remaining
grams NaOH remaining = 0.046 moles NaOH x 40 g/mole = 1.84 g NaOH left over
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