Question

How many liters of .500M HCl are needed to react completely with 2.50 liters of lead(ii) nitrate that has a density of 4.53 g/mL

Answer 1

We are given :

• Concentration of HCL = ,0.500 M

,

• Volume of lead(ii) nitrate = 2.5 L = 2.5 *1000 = ,2500 ml

,

• Density of leadii) nitrate =, 4.53g/L

,

• Volume of HCL =?

1. Calculate mass of lead(ii) nitrate :

$\begin{gathered} \rho\text{ = }\frac{mass\text{ }}{\text{volume }} \\ \therefore\text{ mass = }\rho\text{ X volume } \\ \text{ = 4.53g/ml }\cdot\text{ }2500\text{ ml } \\ \text{ =11325 grams } \end{gathered}$

2. Calculate moles of lead (ii) nitrate :

• Molecular mass of lead(ii) nitrate = 331,2 g/mol

$\begin{gathered} \text{moles = }\frac{mass\text{ }}{\text{molecular mass }} \\ \text{ =}\frac{11325g}{331.2g}\cdot\text{ mol} \\ =\text{ 34.18 moles } \end{gathered}$

3. Calculate moles of HCL

Soichemistry chemical equation shows that :

$\text{HCl +Pb(NO}_3)_2\Rightarrow PbCl_2+HNO_3$

• meaning : 1 mole of HCL reacts with 1 mole of Pb(NO3)2

,

• Therefore 34.18 moles HCL will react with 34.18 moles Pb(No3)2

,

• number of moles of HCL = 34.18

4. Calculate volume of HCl needed

• Concentration HCl = 0.5 M and molesHCl = 34.18

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