Answer:
a) Mg (OH)2
b) 1.12g (approximate)
Step-by-step explanation:
(a) To determine the limiting reagent, we need to compare the moles of Mg(OH)2 and HCl. First calculate the molar mass of Mg(OH)2:
Mg(OH)2 molar mass = mass / molar mass
Mg(OH)2 molar mass = 24.31 g/mol + 2 * (1.01 g/ mol + 16.00) ) g / mol) = 58.32 g / mol
Mg (OH) 2 mol = 1.12 g / 58.32 g / mol ≈ 0.0192 mol
Now, calculate the moles of HCl:
HCl mol = volume * HCl2 mol = 40 L concentration 444
1701 mol/L = 0.0401 mol
When we compare the moles, we see that Mg (OH) 2 is the limiting reagent as it has fewer moles (0.0192 mol) compared to HCl (0.0401 mol).
(b) The balanced equation shows that 1 mole of Mg(OH)2 produces 1 mole of MgCl2.
Thus, the amount of magnesium chloride produced will be equal to the Mg(OH)2 used. Thus, approximately 1.12 g of magnesium chloride is produced.