Write a balanced equation for the double-replacement precipitation reaction described, using the smallest possible integer coefficients.
A precipitate forms when aqueous solutions of iron(II) iodide and sodium hydroxide are combined.
FeI2 (aq) + 2NaOH (aq) ==>Fe(OH)2(s) +2NaI (aq)
To write a balanced equation for a double-replacement precipitation reaction, we need to identify the reactants and products involved.
The reactants are iron(II) iodide (FeI2) and sodium hydroxide (NaOH). Iron(II) iodide is a compound made up of iron (Fe) and iodine (I), while sodium hydroxide is made up of sodium (Na) and hydroxide ion (OH-).
The products of this reaction will be a precipitate, which is formed when two ions switch places. In this case, the iron(II) ion (Fe2+) from the iron(II) iodide will combine with the hydroxide ion (OH-) from the sodium hydroxide, resulting in the formation of iron(II) hydroxide (Fe(OH)2). The other product will be sodium iodide (NaI), formed when sodium (Na+) and iodide (I-) ions combine.
Now, let's write the balanced equation:
FeI2 + 2NaOH → Fe(OH)2 + 2NaI
In this equation, the coefficients in front of each compound show the ratio of reactants and products needed for a balanced equation. The smallest possible integer coefficients are used to maintain the ratio of atoms on both sides of the equation.
Thus, the balanced equation for the double-replacement precipitation reaction involving iron(II) iodide and sodium hydroxide is:
FeI2 + 2NaOH → Fe(OH)2 + 2NaI