Silicon is produced for the chemical and electronics industries by the following reactions. Give the balanced equation for each reaction. a. \(\mathrm{SiO}_{2}(s)+\mathrm{C}(s)\) $\frac{\text { Electric }}{\text { are furnace }}\( \)\mathrm{Si}(s)+\mathrm{CO}(g)$ b. Liquid silicon tetrachloride is reacted with very pure solid magnesium, producing solid silicon and solid magnesium chloride. c. $\mathrm{Na}_{2} \mathrm{SiF}_{6}(s)+\mathrm{Na}(s) \rightarrow \mathrm{Si}(s)+\mathrm{NaF}(s)$

The given reaction is: \(\mathrm{SiO}_{2}(s)+\mathrm{C}(s) \rightarrow \mathrm{Si}(s)+\mathrm{CO}(g)\) To balance this equation, we need to balance the number of atoms of each element on both sides. The equation is already balanced as there are equal numbers of silicon, oxygen, and carbon atoms on both sides of the equation. So, the balanced equation for reaction a is: \(\mathrm{SiO}_{2}(s)+\mathrm{C}(s) \rightarrow \mathrm{Si}(s)+\mathrm{CO}(g)\)

Reaction b is described as the reaction between liquid silicon tetrachloride and solid magnesium. The products are solid silicon and solid magnesium chloride. We can write this reaction as: \(\mathrm{SiCl}_{4}(l) + \mathrm{Mg}(s) \rightarrow \mathrm{Si}(s) + \mathrm{MgCl}_{2}(s)\) Now we need to balance the equation: Both silicon and magnesium are already balanced, but we need to balance the number of chlorine atoms. We can do this by placing a 2 in front of the magnesium and magnesium chloride to balance the atoms: \(\mathrm{SiCl}_{4}(l) + 2\mathrm{Mg}(s) \rightarrow \mathrm{Si}(s) + 2\mathrm{MgCl}_{2}(s)\) The balanced equation for reaction b is: \(\mathrm{SiCl}_{4}(l) + 2\mathrm{Mg}(s) \rightarrow \mathrm{Si}(s) + 2\mathrm{MgCl}_{2}(s)\)

The given reaction is: \(\mathrm{Na}_{2} \mathrm{SiF}_{6}(s)+\mathrm{Na}(s) \rightarrow \mathrm{Si}(s)+\mathrm{NaF}(s)\) To balance this equation, we need to balance the number of atoms of each element on both sides. First, let's balance the sodium atoms by placing a 2 in front of the sodium fluoride: \(\mathrm{Na}_{2} \mathrm{SiF}_{6}(s)+\mathrm{Na}(s) \rightarrow \mathrm{Si}(s)+ 2\mathrm{NaF}(s)\) Now we balance the fluorine atoms by placing a 3 in front of the sodium fluoride: \(\mathrm{Na}_{2} \mathrm{SiF}_{6}(s)+\mathrm{Na}(s) \rightarrow \mathrm{Si}(s)+ 3\mathrm{NaF}(s)\) Finally, balance the sodium atoms by placing a 3 in front of solid sodium: \(\mathrm{Na}_{2} \mathrm{SiF}_{6}(s)+ 3\mathrm{Na}(s) \rightarrow \mathrm{Si}(s)+ 3\mathrm{NaF}(s)\) The balanced equation for reaction c is: \(\mathrm{Na}_{2} \mathrm{SiF}_{6}(s)+ 3\mathrm{Na}(s) \rightarrow \mathrm{Si}(s)+ 3\mathrm{NaF}(s)\)