If 16.0 g of Mg (OH)2 and 11.0 g of HCI are combined...what is the excess reactant?
Mg(OH)2 + 2HCl = MgCl2 + 2H2O
For the reaction with 1 mol Mg(OH)2 0.5 mol HCl is needed
n = "\\frac{m}{M}"
n(Mg(OH)2) = "\\frac{16.0}{58,4} = 0.274 (mol)"
n(HCl) = "\\frac{11}{36.5} = 0.31 (mol)"
"\\frac{0.31}{2} = 0.165"
Answer: Mg(OH)2 is an excess reactant.
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