Question

Several brands of antacids use Al(OH)3 to react with stomach acid, which contains primarily HCl: αAl(OH)3(s)+βHCl(aq)→γAlCl3(aq)+δH2O(l)

A.Calculate the number of grams of AlCl3 formed when 0.430 g of Al(OH)3 reacts.

B. Calculate the number of grams of H2O formed when 0.430 g of Al(OH)3 reacts.

Please show all the steps. I think I'm missing a step.Thanks so much!

Answer 1

A)

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(Al(OH)3)= 0.43 g

number of mol of Al(OH)3,

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

=(0.43 g)/(78.004 g/mol)

= 5.513*10^-3 mol

Balanced chemical equation is:

Al(OH)3 + 3 HCl ---> AlCl3 + 3 H2O

Molar mass of AlCl3,

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

= 1*26.98 + 3*35.45

= 133.33 g/mol

According to balanced equation

mol of AlCl3 formed = (1/1)* moles of Al(OH)3

= (1/1)*0.0055

= 0.0055 mol

mass of AlCl3 = number of mol * molar mass

= 5.513*10^-3*1.333*10^2

= 0.735 g

Answer: 0.735 g

B)

According to balanced equation

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

= (3/1)*0.0055

= 0.0165 mol

mass of H2O = number of mol * molar mass

= 1.654*10^-2*18

= 0.298 g

Answer: 0.298 g