Answer:
D) There are fewer grams of HCl than Mg(OH)₂
Explanation:
The balance chemical equation for given reaction is as;
Mg(OH)₂ + 2 HCl → MgCl₂ + 2 H₂O
Step 1: Calculate Moles of Mg(OH)₂ and HCl as;
Mg(OH)₂:
Moles = Mass / M.Mass
Moles = 25.3 g / 58.319 g/mol
Moles = 0.433 moles of Mg(OH)₂
HCl:
Moles = Mass / M.Mass
Moles = 22.5 g / 36.460 g/mol
Moles = 0.617 moles of HCl
Step 2: Find out Limiting reagent as;
According to equation,
1 mole of Mg(OH)₂ reacted with = 2 moles of HCl
So,
0.433 moles of Mg(OH)₂ will react with = X moles of HCl
Solving for X,
X = 0.433 mol × 2 mol / 1 mol
X = 0.866 moles of HCl
This means for given amount of Mg(OH)₂ we require 0.866 moles of Hcl while, we are only provided with 0.617 moles of HCl hence, HCl is the limiting reagent and will control the final yields of products.
Step 3: Find out Moles of MgCl₂ produced;
According to equation,
2 moles of HCl produced = 1 mole of MgCl₂
So,
0.617 moles of HCl will produce = X moles of MgCl₂
Solving for X,
X = 0.617 mol × 1 mol / 2 mol
X = 0.3085 moles of MgCl₂
Step 4: Calculate Mass of MgCl₂ as;
Mass = Moles × M.Mass
Mass = 0.3085 mol × 36.46 g/mol
Mass = 11.24 g of MgCl₂
Answer:
A) The amounts of both reactants are given.
Explanation:
The amounts of both reactants are given. You cannot tell which reactant will be used up first. Calculate the mass of product that can be produced from the amount of the first reactant; calculate the mass of product that can be produced from the second reactant. Compare the two. The limiting reactant produces the least product.
