2.4 mol Fe(OH)3 & 7.9 mol H2SO4 react according to lớn the equation 2Fe(OH)3+3H2SO4−→ Fe2(SO4)3 +6H2O. If the limiting reactant is Fe(OH)3, determine the amount of excess reactant that remains.
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Looking at the balanced equation:
Since you are told that Fe(OH)3 is the limiting reactant, we will determine how many moles of H2SO4 react và then determine how much H2SO4 is left over at the over of the reaction.
moles H2SO4 used: 2.4 mol Fe(OH)3 x 3 mol H2SO4 / 2 mol Fe(OH)3 = 3.6 moles H2SO4 reacted
moles H2SO4 remaining = initial - used = 7.9 moles - 3.6 moles = 4.3 moles H2SO4 remaining
mass H2SO4 remaining = 4.3 moles H2SO4 x 98 g/mol = 421 g H2SO4 remaining
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