Answer:
1) 11.64 mol/L is the molarity of concentrated HCl.
2) 135.40 mL of volume of 11.64 M will need to prepare 985 mL of 1.6 M HCl.
3) 1,711.08 grams of sodium bicarbonate would be needed to neutralize the spill HCl solution.
Explanation:
HCl solution with 36.0% HCl by mass, menas that in 100 g of solution 36.0 gram of HCl is present.
Mass of HCl= 36.0 g
Moles of HCl = ![\frac{36.0 g}{36.5 g/mol}=0.9863 mol](https://tex.z-dn.net/?f=%5Cfrac%7B36.0%20g%7D%7B36.5%20g%2Fmol%7D%3D0.9863%20mol)
Mass of solution ,m= 100 g
Volume of solution = V = ?
Density of the solution ,d= 1.18 g/mL
![V=\frac{d}{M}=\frac{100 g}{1.18 g/mL}=84.75 mL=0.08475 L](https://tex.z-dn.net/?f=V%3D%5Cfrac%7Bd%7D%7BM%7D%3D%5Cfrac%7B100%20g%7D%7B1.18%20g%2FmL%7D%3D84.75%20mL%3D0.08475%20L)
![Molarity=\frac{Moles }{\text{Volume of solution (L)}}](https://tex.z-dn.net/?f=Molarity%3D%5Cfrac%7BMoles%20%7D%7B%5Ctext%7BVolume%20of%20solution%20%28L%29%7D%7D)
Molarity of the solution :
![=\frac{0.9863 mol}{0.8475 L}=11.64 mol/L](https://tex.z-dn.net/?f=%3D%5Cfrac%7B0.9863%20mol%7D%7B0.8475%20L%7D%3D11.64%20mol%2FL)
11.64 mol/L is the molarity of concentrated HCl.
2)
( Dilution equation)
![M_1= 11.64 M](https://tex.z-dn.net/?f=M_1%3D%2011.64%20M)
![V_1=?](https://tex.z-dn.net/?f=V_1%3D%3F)
![M_2=1.6 M](https://tex.z-dn.net/?f=M_2%3D1.6%20M)
![V_2=985 mL](https://tex.z-dn.net/?f=V_2%3D985%20mL)
![V_1=\frac{M_2V_2}{M_1}=\frac{1.6 M\times 985 mL}{11.64 M}](https://tex.z-dn.net/?f=V_1%3D%5Cfrac%7BM_2V_2%7D%7BM_1%7D%3D%5Cfrac%7B1.6%20M%5Ctimes%20985%20mL%7D%7B11.64%20M%7D)
![=135.40 mL](https://tex.z-dn.net/?f=%3D135.40%20mL)
135.40 mL of volume of 11.64 M will need to prepare 985 mL of 1.6 M HCl.
3.
![NaHCO_3+HCl\rightarrow NaCl+H_2O+CO_2](https://tex.z-dn.net/?f=NaHCO_3%2BHCl%5Crightarrow%20NaCl%2BH_2O%2BCO_2)
Concentration of HCl solution = 11.64 M
Volume of the HCl solution = 1.75 L
Moles of HCl in 1.75 L solution = n
![11.64 M=\frac{n}{1.75 L}](https://tex.z-dn.net/?f=11.64%20M%3D%5Cfrac%7Bn%7D%7B1.75%20L%7D)
![n=1.75 L\times 11.64 M=20.37 mol](https://tex.z-dn.net/?f=n%3D1.75%20L%5Ctimes%2011.64%20M%3D20.37%20mol)
According to reaction 1 mole of HCl neutralized by 1 mole of sodium carbonate.
Then 20.37 moles of HCl will neutralized by ;
of sodium carbonate
Mass of 20.37 moles of sodium carbonate :
= 20.37 mol × 84g/mol = 1,711.08 g
1,711.08 grams of sodium bicarbonate would be needed to neutralize the spill HCl solution.