Question

barium chloride and sodium sulfate react according to the following equation.
bacl₂ + na₂so₄ → baso₄ + 2nacl
answer the following question(s) about this reaction.
how many grams of sodium chloride can be produced from 50. grams of barium chloride?
2.0 g
14 g
28 g
59 g
208 g

Explanation:

Step1: Calculate moles of \(BaCl_2\)

Molar mass of \(BaCl_2\) (\(M_{BaCl_2}\)): \(M_{Ba}=137.33\space g/mol\), \(M_{Cl}=35.45\space g/mol\), so \(M_{BaCl_2}=137.33+(2\times35.45)=208.23\space g/mol\).
Moles of \(BaCl_2\) (\(n_{BaCl_2}\)): \(n_{BaCl_2}=\frac{m_{BaCl_2}}{M_{BaCl_2}}=\frac{50\space g}{208.23\space g/mol}\approx0.24\space mol\)

Step2: Use mole ratio from balanced equation

From \(BaCl_2 + Na_2SO_4
ightarrow BaSO_4 + 2NaCl\), mole ratio \(n_{BaCl_2}:n_{NaCl}=1:2\).
So moles of \(NaCl\) (\(n_{NaCl}\)): \(n_{NaCl}=2\times n_{BaCl_2}=2\times0.24 = 0.48\space mol\)

Step3: Calculate mass of \(NaCl\)

Molar mass of \(NaCl\) (\(M_{NaCl}\)): \(M_{Na}=22.99\space g/mol\), \(M_{Cl}=35.45\space g/mol\), so \(M_{NaCl}=22.99 + 35.45=58.44\space g/mol\)
Mass of \(NaCl\) (\(m_{NaCl}\)): \(m_{NaCl}=n_{NaCl}\times M_{NaCl}=0.48\space mol\times58.44\space g/mol\approx28\space g\)

Answer:

28 g