Final answer:
To find the mass of magnesium nitrate produced from 8.00 mol of nitric acid, the number of moles of magnesium nitrate is calculated using stoichiometry and then converted to grams. The correct answer is c. 593 g of magnesium nitrate.
Step-by-step explanation:
To determine how many grams of magnesium nitrate are produced from 8.00 mol of nitric acid, HNO3, given the reaction 2HNO3 + Mg(OH)2 → Mg(NO3)2 + 2H2O, we need to use stoichiometry.
The reaction shows a molar ratio of 2 moles of HNO3 to 1 mole of Mg(NO3)2. Since we have an excess of Mg(OH)2, nitric acid is the limiting reactant. We can divide the amount of HNO3 used by the molar ratio to find the moles of Mg(NO3)2 produced:
8.00 mol HNO3 × (1 mol Mg(NO3)2 / 2 mol HNO3) = 4.00 mol Mg(NO3)2
To convert moles of magnesium nitrate to grams, we use its molar mass (148.31 g/mol):
4.00 mol Mg(NO3)2 × 148.31 g/mol = 593 g of Mg(NO3)2
Therefore, the correct answer is c. 593 g of magnesium nitrate.