Respuesta :
Answer:
a. The maximum volume of 0.143 M HCl required is 154.4 mL.
b. The maximum volume of 0.143 M HCl required is 135.7 mL.
Explanation:
a.
[tex]Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O[/tex]
Mass of aluminum hydroxide = 350 mg = 0.350 g ( 1mg = 0.001 g)
Moles of aluminum hydroxide = [tex]\frac{0.350 g}{78 g/mol}=0.004487 mol[/tex]
According to reaction ,3 moles of HCl neutralize 1 mole of aluminum hydroxide.Then 0.004487 mole of aluminum hydroxide will be neutralize by :
[tex]\frac{3}{1}\times 0.004487 mol=0.01346 mol[/tex] of HCl.
[tex]Mg(OH)_2+2HCl\rightarrow MgCL_2+2H_2O[/tex]
Mass of magnesium hydroxide = 250 mg = 0.250 g ( 1mg = 0.001 g)
Moles of magnesium hydroxide = [tex]\frac{0.250 g}{58 g/mol}=0.004310 mol[/tex]
According to reaction ,2 moles of HCl neutralize 1 mole of magnesium hydroxide.Then 0.004310 mole of magnesium hydroxide will be neutralize by :
[tex]\frac{2}{1}\times 0.004310 mol=0.008621 mol[/tex] of HCl.
Total moles of HCl required to neutralize both :
0.01346 mol + 0.008621 mol = 0.02208 mol
Molarity of the HCL solution = 0.143 M
Volume of the solution = V
[tex]Molarity=\frac{\text{Total moles of HCl}{\text{Volume in Liter}}[/tex]
[tex]V=\frac{0.02208 mol}{0.143 M}=0.1544 L[/tex]
1 L = 1000 mL
0.1544 L = 154.4 mL
The maximum volume of 0.143 M HCl required is 154.4 mL.
b.
[tex]CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2[/tex]
Mass of calcium carbonate = 970mg = 0.970 g ( 1mg = 0.001 g)
Moles of calcium carbonate = [tex]\frac{0.970 g}{100 g/mol}=0.00970 mol[/tex]
According to reaction ,2 moles of HCl neutralize 1 mole of calcium carbonate.Then 0.00970 mole of calcium carbonate will be neutralize by :
[tex]\frac{2}{1}\times 0.00970 mol=0.0194 mol[/tex] of HCl.
Total moles of HCl required to neutralize calcium carbonate : 0.0194 mol
Molarity of the HCL solution = 0.143 M
Volume of the solution = V
[tex]Molarity=\frac{\text{Total moles of HCl}}{\text{Volume in Liter}}[/tex]
[tex]V=\frac{0.0194 mol}{0.143 M}=0.1357 L[/tex]
1 L = 1000 mL
0.1357 L = 135.7 mL
The maximum volume of 0.143 M HCl required is 135.7 mL.
