Question

what mass of nacl formed when 0.25 g na react completely with 0.39 g cl2? 0.59 g nacl -0.14 g nacl 0.64 g nacl 0.14 g nacl

Explanation:

Step1: Write the balanced chemical equation

$$2Na + Cl_{2}=2NaCl$$

Step2: Calculate the molar masses

The molar mass of \(Na\) is \(M_{Na}=23\space g/mol\), the molar mass of \(Cl_{2}\) is \(M_{Cl_{2}} = 71\space g/mol\), and the molar mass of \(NaCl\) is \(M_{NaCl}=58.5\space g/mol\)

Step3: Calculate the moles of reactants

The moles of \(Na\), \(n_{Na}=\frac{m_{Na}}{M_{Na}}=\frac{0.25\space g}{23\space g/mol}\approx0.0109\space mol\)
The moles of \(Cl_{2}\), \(n_{Cl_{2}}=\frac{m_{Cl_{2}}}{M_{Cl_{2}}}=\frac{0.39\space g}{71\space g/mol}\approx0.0055\space mol\)

Step4: Determine the limiting reactant

From the balanced equation \(2Na + Cl_{2}=2NaCl\), the mole ratio of \(Na\) to \(Cl_{2}\) is \(2:1\).
For \(n_{Na} = 0.0109\space mol\), the required \(n_{Cl_{2}}\) is \(\frac{0.0109}{2}=0.00545\space mol\). Since \(0.00545\space mol<0.0055\space mol\), \(Na\) is the limiting reactant.

Step5: Calculate the moles of \(NaCl\) formed

From the balanced equation, \(n_{NaCl}=n_{Na}\) (because of the mole ratio \(2:2 = 1:1\) for \(Na\) and \(NaCl\))

Step6: Calculate the mass of \(NaCl\)

\(m_{NaCl}=n_{NaCl}\times M_{NaCl}\)
Since \(n_{NaCl}=n_{Na}\approx0.0109\space mol\), \(m_{NaCl}=0.0109\space mol\times58.5\space g/mol\approx0.64\space g\)

Answer:

\(0.64\space g\space NaCl\) (corresponding to the third option)