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:

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

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}=\frac{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.