Question

Using the reaction below, how many molecules of lithium nitrate (LiNo3) will be needed to react with 2.35 grams of lead (IV) sulfate, Pb (SO4)2?equation in photo

Answer 1

From the reaction, each Pb(SO₄)₂ reacts with 4 LiNO₃.

So, first, we need to figure ot how many moles of Pb(SO₄)₂ we have in 2.35 grams.

For this, we need to use the molar masses of its elements to find its molar mass:

`\begin{gathered} M_(Pb\mleft(SO_4\mright)_2)=1\cdot M_(Pb)+2\cdot(1\cdot M_S+4\cdot M_O)=(1\cdot207.2+2\cdot(1\cdot32.065+4\cdot15.9994))g\/mol \\ M_(Pb(SO_4)_2)=399.3252g\/mol \end{gathered}`

Using it, we can calcualte the number of moles of Pb(SO₄)₂:

`\begin{gathered} M_(Pb\mleft(SO_4\mright)_2)=\frac{m_(Pb\mleft(SO_4\mright)_2)}{n_{Pb\mleft(SO_(4)\mright)_(2)}} \\ n_(Pb\mleft(SO_4\mright)_2)=\frac{m_(Pb\mleft(SO_4\mright)_2)}{M_{Pb\mleft(SO_(4)\mright)_(2)}}=(2.35g)/(399.3252g\/mol)=0.005884\ldots mol \end{gathered}`

Using the stoichiometry, we have:

LiNO₃ --- Pb(SO₄)₂

4 --- 1

`\begin{gathered} (n_(LiNO_3))/(4)=(n_(Pb\mleft(SO_4\mright)_2))/(1) \\ n_(LiNO_3)=4\cdot n_(Pb\mleft(SO_4\mright)_2)=4\cdot0.005884\ldots mol=0.023539\ldots mol \end{gathered}`

Now that we know the number of moles of LiNO₃, we can calculate how many LiNO₃ are needed by using the Avogadro's Number:

`N_(LiNO_3)=N_A\cdot n_(LiNO_3)`

Avogadro's Number is approximately 6.022 x 10²³ mol⁻¹, so, using this value, we have:

`N_(LiNO_3)=6.022*10^(23)mol^(-1)\cdot0.023539\ldots mol=1.41756\ldots*10^(22)\approx1.42*10^(22)`

So, we need approximately 1.42 x 10²² LiNO₃.