Question

Al(OH)3(s)+3HCl(aq)→AlCl3(aq)+3H2O(l)

1. Calculate the number of grams of HCl that can react with 0.440 g of Al(OH)3.

2. Calculate the number of grams of AlCl3 formed when 0.440 g of Al(OH)3 reacts.

3. Calculate the number of grams of H2O formed when 0.440 g of Al(OH)3 reacts.

Answer 1

1)

Molar mass of Al(OH)3,

MM = 1*MM(Al) + 3*MM(O) + 3*MM(H)

= 1*26.98 + 3*16.0 + 3*1.008

= 78.004 g/mol

mass of Al(OH)3 = 0.44 g

mol of Al(OH)3 = (mass)/(molar mass)

= 0.44/78

= 5.641*10^-3 mol

Balanced chemical equation is:

Al(OH)3(s)+3HCl(aq)→AlCl3(aq)+3H2O(l)

According to balanced equation

mol of HCl reacted = (3/1)* moles of Al(OH)3

= (3/1)*5.641*10^-3

= 1.692*10^-2 mol

Molar mass of HCl,

MM = 1*MM(H) + 1*MM(Cl)

= 1*1.008 + 1*35.45

= 36.458 g/mol

mass of HCl = number of mol * molar mass

= 1.692*10^-2*36.46

= 0.6169 g

Answer: 0.617 g

2)

According to balanced equation

mol of AlCl3 formed = moles of Al(OH)3

= 5.641*10^-3 mol

Molar mass of AlCl3,

MM = 1*MM(Al) + 3*MM(Cl)

= 1*26.98 + 3*35.45

= 133.33 g/mol

mass of AlCl3 = number of mol * molar mass

= 5.641*10^-3*1.333*10^2

= 0.7521 g

Answer: 0.752 g

3)

According to balanced equation

mol of H2O formed = (3/1)* moles of Al(OH)3

= (3/1)*5.641*10^-3

= 1.692*10^-2 mol

Molar mass of H2O,

MM = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

mass of H2O = number of mol * molar mass

= 1.692*10^-2*18.02

= 0.3049 g

Answer: 0.305 g