Final answer:
The mass of solid formed when 54.0 ml of 3.70 M sodium hydroxide is combined with 44.0 ml of 1.30 M magnesium chloride is 3.34 grams.
Step-by-step explanation:
To find the mass of solid formed when 54.0 ml of 3.70 M sodium hydroxide (NaOH) is combined with 44.0 ml of 1.30 M magnesium chloride (MgCl2), we need to determine the limiting reagent in the reaction and use stoichiometry to calculate the mass of the solid formed.
The balanced equation for the reaction is:
MgCl₂ + 2NaOH -> Mg(OH)₂ + 2NaCl
By comparing the number of moles of NaOH and MgCl2, we find that:
1 mole of NaOH reacts with 1 mole of MgCl2
Thus, the limiting reagent is determined by the amount of MgCl2 present. We can calculate the moles of MgCl2 using the given concentration and volume:
Moles of MgCl2 = concentration x volume = 1.30 M x (44.0/1000) L = 0.0572 moles
The molar ratio between MgCl2 and Mg(OH)₂ is 1:1. Therefore, the number of moles of Mg(OH)₂ formed is also 0.0572 moles.
The molar mass of Mg(OH)₂ is 58.33 g/mol. Therefore, the mass of the solid formed is:
Mass = moles x molar mass = 0.0572 moles x 58.33 g/mol = 3.34 g
Therefore, the correct answer is A) 9.84 g.