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balance the reactions and state the type of reaction described. 1. sodi…

Question

balance the reactions and state the type of reaction described.

  1. sodium metal reacts vigorously with water giving off a gas.
  2. hydrogen chloride gas is commercially made by reacting table salt with sulfuric acid.
  3. molten iron produced by the highly exothermic thermite reaction was used to weld railroad rails.

Explanation:

Step1: Analyze the first reaction

Sodium (\(Na\)) reacts with water (\(H_2O\)). The un - balanced equation is \(Na(s)+H_2O(l)
ightarrow H_2(g)+NaOH(aq)\).
For sodium: On the left - hand side, there is 1 \(Na\) atom, and on the right - hand side, there is 1 \(Na\) atom in \(NaOH\).
For hydrogen: On the left - hand side, there are 2 \(H\) atoms in \(H_2O\), and on the right - hand side, there are 3 \(H\) atoms (\(2\) in \(H_2\) and \(1\) in \(NaOH\)).
For oxygen: On the left - hand side, there is 1 \(O\) atom in \(H_2O\), and on the right - hand side, there is 1 \(O\) atom in \(NaOH\).
We start by balancing hydrogen. If we put a coefficient of \(2\) in front of \(Na\) and \(2\) in front of \(H_2O\) and \(2\) in front of \(NaOH\), the equation becomes \(2Na(s)+2H_2O(l)
ightarrow H_2(g)+2NaOH(aq)\). Now, for sodium: \(2\) on the left and \(2\) on the right. For hydrogen: \(4\) on the left (\(2\times2\) from \(H_2O\)) and \(4\) on the right (\(2\) from \(H_2\) and \(2\) from \(2NaOH\)). For oxygen: \(2\) on the left (\(2\times1\) from \(H_2O\)) and \(2\) on the right (\(2\times1\) from \(2NaOH\)).

Step2: Analyze the second reaction

Hydrogen chloride (\(HCl\)) is made by reacting table salt (\(NaCl\)) with sulfuric acid (\(H_2SO_4\)). The un - balanced equation is \(NaCl(s)+H_2SO_4(aq)
ightarrow HCl(g)+Na_2SO_4(aq)\).
For sodium: \(1\) on the left and \(2\) on the right.
For chlorine: \(1\) on the left and \(1\) on the right.
For hydrogen: \(2\) on the left and \(1\) on the right.
For sulfur: \(1\) on the left and \(1\) on the right.
For oxygen: \(4\) on the left and \(4\) on the right.
We put a coefficient of \(2\) in front of \(NaCl\) and \(2\) in front of \(HCl\). The balanced equation is \(2NaCl(s)+H_2SO_4(aq)
ightarrow 2HCl(g)+Na_2SO_4(aq)\)

Step3: Analyze the third reaction

The thermite reaction \(Al + Fe_2O_3
ightarrow Fe+Al_2O_3\).
For aluminum: \(1\) on the left and \(2\) on the right.
For iron: \(2\) on the left and \(1\) on the right.
For oxygen: \(3\) on the left and \(3\) on the right.
We put a coefficient of \(2\) in front of \(Al\) and \(2\) in front of \(Fe\). The balanced equation is \(2Al(s)+Fe_2O_3(s)
ightarrow 2Fe(l)+Al_2O_3(s)\)

Answer:

  1. \(2Na(s)+2H_2O(l)

ightarrow H_2(g)+2NaOH(aq)\)

  1. \(2NaCl(s)+H_2SO_4(aq)

ightarrow 2HCl(g)+Na_2SO_4(aq)\)

  1. \(2Al(s)+Fe_2O_3(s)

ightarrow 2Fe(l)+Al_2O_3(s)\)