Question

A 50.6g sample of Mg(OH)2 is redonaldtrumpacted with 45.0g of HCl according to the reaction (already balanced): Mg(OH)2 + 2 HCl --> MgCl2 + 2 H2O Here are the specific questions: How much excess Mg(OH)2 remains after the HCl is depleted? What is the theoretical yield of MgCl2?

Answer 1

Molar mass of Mg(OH)2,
MM = 1*MM(Mg) + 2*MM(O) + 2*MM(H)
= 1*24.31 + 2*16.0 + 2*1.008
= 58.326 g/mol


mass(Mg(OH)2)= 50.6 g

number of mol of Mg(OH)2,
n = mass of Mg(OH)2/molar mass of Mg(OH)2
=(50.6 g)/(58.326 g/mol)
= 0.8675 mol

Molar mass of HCl,
MM = 1*MM(H) + 1*MM(Cl)
= 1*1.008 + 1*35.45
= 36.458 g/mol


mass(HCl)= 45.0 g

number of mol of HCl,
n = mass of HCl/molar mass of HCl
=(45.0 g)/(36.458 g/mol)
= 1.234 mol
Balanced chemical equation is:
Mg(OH)2 + 2 HCl ---> MgCl2 + 2 H2O


1 mol of Mg(OH)2 reacts with 2 mol of HCl
for 0.867538 mol of Mg(OH)2, 1.735075 mol of HCl is required
But we have 1.234297 mol of HCl

so, HCl is limiting reagent
we will use HCl in further calculation


1)
According to balanced equation
mol of Mg(OH)2 reacted = (1/2)* moles of HCl
= (1/2)*1.234297
= 0.617148 mol
mol of Mg(OH)2 remaining = mol initially present - mol reacted
mol of Mg(OH)2 remaining = 0.867538 - 0.617148
mol of Mg(OH)2 remaining = 0.250389 mol


Molar mass of Mg(OH)2,
MM = 1*MM(Mg) + 2*MM(O) + 2*MM(H)
= 1*24.31 + 2*16.0 + 2*1.008
= 58.326 g/mol

mass of Mg(OH)2,
m = number of mol * molar mass
= 0.2504 mol * 58.326 g/mol
= 14.6 g

Answer: 14.6 g of Mg(OH)2 remains

2)
Molar mass of MgCl2,
MM = 1*MM(Mg) + 2*MM(Cl)
= 1*24.31 + 2*35.45
= 95.21 g/mol

According to balanced equation
mol of MgCl2 formed = (1/2)* moles of HCl
= (1/2)*1.234297
= 0.617148 mol


mass of MgCl2 = number of mol * molar mass
= 0.6171*95.21
= 58.8 g
Answer: 58.8 g of MgCl2 formed