Question

Na2CO3(s) + 2HCl(aq) -->2NaCl(aq) + CO2(g) + H2O(l)What mass of sodium chloride can be produced from 27.2 g of sodium carbonate?

Answer 1

The sodium chloride is NaCl in the equation, while the sodium carbonate is the Na₂CO₃.

The stoichiometry of them in this reaction is 1 Na₂CO₃ to 2 NaCl.

So, first we need to convert the 27.2g of Na₂CO₃ to number of moles of Na₂CO₃. We do that by using the molar mass of Na₂CO₃:

`\begin{gathered} M_{Na_(2)CO_(3)}=\frac{m_(Na_2CO_3)}{n_{Na_(2)CO_(3)}} \\ n_{Na_(2)CO_(3)}=\frac{m_(Na_2CO_3)}{M_{Na_(2)CO_(3)}} \end{gathered}`

The molar mass of Na₂CO₃ can be calculated using the molar masses of Na, C and O, which can be consulted in a periodic table:

`\begin{gathered} M_(Na)\approx22.9898g/mol \\ M_C\approx12.0107g/mol \\ M_O\approx15.9994g/mol \end{gathered}`
`\begin{gathered} M_(Na_2CO_3)=2\cdot M_(Na)+1\cdot M_C+3\cdot M_O \\ M_(Na_2CO_3)\approx(2\cdot22.9898+1\cdot12.0107+3\cdot15.9994)g/mol \\ M_(Na_2CO_3)\approx105.9885g/mol \end{gathered}`

So, the number of moles of Na₂CO₃ is:

`n_{Na_(2)CO_(3)}=\frac{m_{Na_(2)CO_(3)}}{M_{Na_(2)CO_(3)}}\approx(27.2g)/(105.9885g/mol)\approx0.2566mol`

Now, since the stoichimetry is 1 Na₂CO₃ to 2 NaCl, each mol of Na₂CO₃ will produce 2 moles of NaCl, thus:

`n_(NaCl)=2\cdot n_(Na_2CO_3)\approx2\cdot0.2566mol=0.5132mol`

Now, we use the molar mass of NaCl to calculate the mass of it:

`\begin{gathered} M_(Na)\approx22.9898g/mol \\ M_(Cl)\approx35.453g/mol \end{gathered}`

`\begin{gathered} M_(NaCl)=1\cdot M_(Na)+1\cdot M_(Cl) \\ M_(NaCl)\approx(1\cdot22.9898+1\cdot35.453)g/mol \\ M_(NaCl)\approx58.4428g/mol \end{gathered}`
`\begin{gathered} M_(NaCl)=(m_(NaCl))/(n_(NaCl)) \\ m_(NaCl)=n_(NaCl)\cdot M_(NaCl)\approx0.5132mol\cdot58.4428g/mol=29.99\ldots g\approx30.0g \end{gathered}`

So, the mass of sodium chloride that can be produced from 27.2g of sodium carbonate is approximately 30.0g.