Silicon is produced for the chemical and electronics industries by the following reactions. Give the balanced equation for each reaction. a. $\operatorname{Si} \mathrm{O}_{2}(s)+\mathrm{C}(s) \longrightarrow \mathrm{Si}(s)+\mathrm{CO}(g)$ b. Silicon tetrachloride is reacted with very pure magne- sium, producing silicon and magnesium chloride. c. $\mathrm{Na}_{2} \mathrm{SiF}_{6}(s)+\mathrm{Na}(s) \longrightarrow \mathrm{Si (s)+\mathrm{NaF}(s)$

Given the equation: \(\mathrm{SiO_2}(s) + \mathrm{C}(s) \longrightarrow \mathrm{Si}(s) + \mathrm{CO}(g)\) This equation is already balanced, and there is no need to add any coefficients. Balanced equation: \(\mathrm{SiO_2}(s) + \mathrm{C}(s) \longrightarrow \mathrm{Si}(s) + \mathrm{CO}(g)\)

Given the information: Silicon tetrachloride (SiCl₄) reacts with very pure magnesium (Mg) to produce silicon (Si) and magnesium chloride (MgCl₂). First, write the unbalanced equation: \(\mathrm{SiCl_4} + \mathrm{Mg} \longrightarrow \mathrm{Si} + \mathrm{MgCl_2}\) Now, we need to balance it. There is one Si on both sides, but four Cl on the left side and two Cl on the right side. To fix this, add a coefficient of 2 in front of MgCl₂ on the right. So, the equation becomes: \(\mathrm{SiCl_4} + \mathrm{Mg} \longrightarrow \mathrm{Si} + 2\mathrm{MgCl_2}\) Now there are four Cl on both sides, but there are two Mg on the right side and only one Mg on the left side. To fix this, add a coefficient of 2 in front of Mg on the left. Balanced equation: \(\mathrm{SiCl_4} + 2\mathrm{Mg} \longrightarrow \mathrm{Si} + 2\mathrm{MgCl_2}\)

Given the equation: \(\mathrm{Na_2SiF_6}(s) + \mathrm{Na}(s) \longrightarrow \mathrm{Si}(s) + \mathrm{NaF}(s)\) First, balance the Si atoms. There is already one Si atom on both sides, so it is balanced. Next, balance the F atoms. There are 6 F atoms on the left side and 1 F atom on the right side. To fix this, add a coefficient of 6 in front of NaF on the right. So, the equation becomes: \(\mathrm{Na_2SiF_6}(s) + \mathrm{Na}(s) \longrightarrow \mathrm{Si}(s) + 6\mathrm{NaF}(s)\) Now, balance the Na atoms. There are 2 Na atoms on the left side and 6 Na atoms on the right side. To fix this, add a coefficient of 3 in front of Na on the left. Balanced equation: \(\mathrm{Na_2SiF_6}(s) + 3\mathrm{Na}(s) \longrightarrow \mathrm{Si}(s) + 6\mathrm{NaF}(s)\)