Final answer:
To find out how many grams of water are needed to react and form 8.20 moles of Mg(OH)2, we use the molar mass of water (18.02 g/mol) and multiply by the number of moles, resulting in 147.764 grams of water.
Step-by-step explanation:
The question 'How many grams of water react to form 8.20 moles of Mg(OH)2?' pertains to stoichiometry, which is a section of chemistry that involves the calculation of reactants and products in chemical reactions. To solve this question, you should first write down the balanced chemical equation for the reaction:
MgO(s) + H2O(l) → Mg(OH)2(s)
Note that according to the balanced chemical equation, one mole of MgO reacts with one mole of water to produce one mole of Mg(OH)2. If we want to produce 8.20 moles of Mg(OH)2, then we need 8.20 moles of water since the molar ratio is 1:1.
Now, using the molar mass of water (18.02 g/mol), we can calculate the mass of water needed:
Mass of water = 8.20 moles × 18.02 g/mol = 147.764 grams of water
147.764 grams of water are required to react and form 8.20 moles of Mg(OH)2.