Answer :
Answer:
750 mg of Al(OH)3 consumes greater mass of HCl
Explanation:
Equation of reaction between CaCO3 and HCl:
CaCO3 + 2HCl = CaCl2 + CO2 + H2O
1 mole (100g) of CaCO3 consumes 2 moles (73g) of HCl
750mg (0.75g) of CaCO3 will consume 0.75×73/100 = 0.5475g of HCl
Equation of reaction between Al(OH)3 and HCl:
Al(OH)3 + 3HCl = AlCl3 + 3H2O
1 mole (78g) of Al(OH)3 consumes 3 mole (109.5g) of HCl
750mg (0.75g) of CaCO3 will consume 0.75×109.5/78 = 1.0529g of HCl
750mg of CaCO3 consumes 0.5475g of HCl while 750mg of Al(OH)3 consumes 1.0529g of HCl
Answer:
AL(OH)3 consumes greter mass of HCl.
Explanation:
Equation of the reaction
CaCO3(s) + HCl(aq) --> CaCl2(aq) + CO2(g) + H2O(g)
Mass of CaCO3 = 750 mg
= 0.75 g
Molar mass = 40 + 12 + (16*3)
= 100 g/mol.
Since 1 mole of CaCO3 reactes with 1 mole of HCl to form 1 mole of CaCl2.
Number of moles = mass/molar mass
= 0.75/100
= 0.0075 moles of CaCO3
= 0.0075 moles of HCl
Molar mass of HCl = 1 + 35.5
= 36.5 g/mol
Mass of HCl consumed = moles*molar mass
= 0.0075*36.5
= 0.27 g of HCl
Equation of reaction
Al(OH)3(aq) + 3HCl(aq) --> AlCl3(aq) + 3H2O(g)
Mass of Al(OH)3 = 750 mg
= 0.75 g
Molar mass = 27 + (1 + 16)*3
= 78 g/mol.
Since 1 mole of Al(OH)3 reactes with 3 mole of HCl to form 1 mole of AlCl3.
Number of moles = mass/molar mass
= 0.75/78
= 0.0096 moles of Al(OH)3
= 0.0096*3
= 0.02885 moles of HCl
Molar mass of HCl = 1 + 35.5
= 36.5 g/mol
Mass of HCl consumed = moles*molar mass
= 0.02885*36.5
= 1.05 g of HCl
Therefore, Al(OH)3 consumes greater mass of HCl.