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If 1.85 g of Mg(OH)2 reacts with 3.71 g of HCl how much MgCl2 is produced
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\[Mg(OH) _{_{2}}+2HCl \rightarrow MgCl _{2}+H_{2}O\] 1 mole of Magnesium hydxide will 2 moles of Hydrochloric acid to give 1 mole of Magnesium Chloride
You have to convert the grams of Mg(OH)2 into moles \[\large 1.85 g Mg(OH)_2 * \frac{1 mol}{58.32 g} = 0.032 mol \] Like Narad stated, it's a 1:1 for Mg(OH)2 and MgCl2, so it would also produce 0.032 mol of MgCl2. Just multiply 0.032 mol by the mass. \[\large 0.032 mol * \frac{95.211 g MgCl_2}{1 mol} = 3.02 g MgCl_2\] So 3.02g of MgCl2 is produced in the reaction.
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