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.
![Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O](https://tex.z-dn.net/?f=Al%28OH%29_3%2B3HCl%5Crightarrow%20AlCl_3%2B3H_2O)
Mass of aluminum hydroxide = 350 mg = 0.350 g ( 1mg = 0.001 g)
Moles of aluminum hydroxide = ![\frac{0.350 g}{78 g/mol}=0.004487 mol](https://tex.z-dn.net/?f=%5Cfrac%7B0.350%20g%7D%7B78%20g%2Fmol%7D%3D0.004487%20mol)
According to reaction ,3 moles of HCl neutralize 1 mole of aluminum hydroxide.Then 0.004487 mole of aluminum hydroxide will be neutralize by :
of HCl.
![Mg(OH)_2+2HCl\rightarrow MgCL_2+2H_2O](https://tex.z-dn.net/?f=Mg%28OH%29_2%2B2HCl%5Crightarrow%20MgCL_2%2B2H_2O)
Mass of magnesium hydroxide = 250 mg = 0.250 g ( 1mg = 0.001 g)
Moles of magnesium hydroxide = ![\frac{0.250 g}{58 g/mol}=0.004310 mol](https://tex.z-dn.net/?f=%5Cfrac%7B0.250%20g%7D%7B58%20g%2Fmol%7D%3D0.004310%20mol)
According to reaction ,2 moles of HCl neutralize 1 mole of magnesium hydroxide.Then 0.004310 mole of magnesium hydroxide will be neutralize by :
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
![Molarity=\frac{\text{Total moles of HCl}{\text{Volume in Liter}}](https://tex.z-dn.net/?f=Molarity%3D%5Cfrac%7B%5Ctext%7BTotal%20moles%20of%20HCl%7D%7B%5Ctext%7BVolume%20in%20Liter%7D%7D)
![V=\frac{0.02208 mol}{0.143 M}=0.1544 L](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B0.02208%20mol%7D%7B0.143%20M%7D%3D0.1544%20L)
1 L = 1000 mL
0.1544 L = 154.4 mL
The maximum volume of 0.143 M HCl required is 154.4 mL.
b.
![CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2](https://tex.z-dn.net/?f=CaCO_3%2B2HCl%5Crightarrow%20CaCl_2%2BH_2O%2BCO_2)
Mass of calcium carbonate = 970mg = 0.970 g ( 1mg = 0.001 g)
Moles of calcium carbonate = ![\frac{0.970 g}{100 g/mol}=0.00970 mol](https://tex.z-dn.net/?f=%5Cfrac%7B0.970%20g%7D%7B100%20g%2Fmol%7D%3D0.00970%20mol)
According to reaction ,2 moles of HCl neutralize 1 mole of calcium carbonate.Then 0.00970 mole of calcium carbonate will be neutralize by :
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
![Molarity=\frac{\text{Total moles of HCl}}{\text{Volume in Liter}}](https://tex.z-dn.net/?f=Molarity%3D%5Cfrac%7B%5Ctext%7BTotal%20moles%20of%20HCl%7D%7D%7B%5Ctext%7BVolume%20in%20Liter%7D%7D)
![V=\frac{0.0194 mol}{0.143 M}=0.1357 L](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B0.0194%20mol%7D%7B0.143%20M%7D%3D0.1357%20L)
1 L = 1000 mL
0.1357 L = 135.7 mL
The maximum volume of 0.143 M HCl required is 135.7 mL.