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 -0.14 g nacl 0.64 g nacl

Explanation:

Step1: Write the chemical equation

The reaction between \(Na\) and \(Cl_2\) is \(2Na + Cl_2=2NaCl\).

Step2: Calculate the moles of \(Na\) and \(Cl_2\)

The molar mass of \(Na\) is \(M_{Na}=23\space g/mol\), so 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 molar mass of \(Cl_2\) is \(M_{Cl_2} = 71\space g/mol\), so the moles of \(Cl_2\), \(n_{Cl_2}=\frac{m_{Cl_2}}{M_{Cl_2}}=\frac{0.39\space g}{71\space g/mol}\approx0.0055\space mol\).
From the 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 \(n_{Cl_2}\) available (\(0.0055\space mol\)) is slightly more than required (\(0.00545\space mol\)), \(Na\) is the limiting reactant.

Step3: Calculate the moles of \(NaCl\)

From the equation \(2Na+Cl_2 = 2NaCl\), the mole ratio of \(Na\) to \(NaCl\) is \(1:1\). So \(n_{NaCl}=n_{Na}\approx0.0109\space mol\).
The molar mass of \(NaCl\) is \(M_{NaCl}=58.5\space g/mol\).
The mass of \(NaCl\), \(m_{NaCl}=n_{NaCl}\times M_{NaCl}=0.0109\space mol\times58.5\space g/mol\approx0.64\space g\).

Answer:

\(0.64\space g\space NaCl\) (the fourth option)