Question

1. add coefficients to balance the equation when aqueous solutions of potassium hydroxide and cobalt(iii) nitrate are mixed.. co(no₃)₃ + koh → co(oh)₃ + kno₃

Explanation:

Step1: Balance Co atoms

There is 1 Co atom on the left (in $\ce{Co(NO_{3})_{3}}$) and 1 Co atom on the right (in $\ce{Co(OH)_{3}}$), so the coefficient for $\ce{Co(NO_{3})_{3}}$ and $\ce{Co(OH)_{3}}$ is 1 for now.

Step2: Balance NO₃⁻ ions

On the left, $\ce{Co(NO_{3})_{3}}$ has 3 $\ce{NO_{3}^-}$ ions. On the right, $\ce{KNO_{3}}$ has 1 $\ce{NO_{3}^-}$ ion. So we need 3 $\ce{KNO_{3}}$ to balance the $\ce{NO_{3}^-}$ ions. So the coefficient for $\ce{KNO_{3}}$ is 3.

Step3: Balance K atoms

Since $\ce{KNO_{3}}$ has a coefficient of 3, there are 3 K atoms on the right. On the left, K is in $\ce{KOH}$, so the coefficient for $\ce{KOH}$ must be 3 to balance K atoms.

Step4: Balance OH⁻ ions

On the left, $\ce{KOH}$ has 3 $\ce{OH^-}$ ions (since coefficient is 3). On the right, $\ce{Co(OH)_{3}}$ has 3 $\ce{OH^-}$ ions (since coefficient of $\ce{Co(OH)_{3}}$ is 1, and each has 3 $\ce{OH^-}$), so OH⁻ is balanced.

Now let's check all atoms:

• Co: 1 (left) and 1 (right) - balanced.
• N: 3 (from $\ce{Co(NO_{3})_{3}}$) and 3 (from 3 $\ce{KNO_{3}}$) - balanced.
• O: Let's count. Left: $\ce{Co(NO_{3})_{3}}$ has 9 O, $\ce{KOH}$ has 3 O (31) → total 12. Right: $\ce{Co(OH)_{3}}$ has 3 O, $\ce{KNO_{3}}$ has 9 O (33) → total 12 - balanced.
• H: Left: $\ce{KOH}$ has 3 H (31). Right: $\ce{Co(OH)_{3}}$ has 3 H (31) - balanced.
• K: 3 (from $\ce{KOH}$) and 3 (from $\ce{KNO_{3}}$) - balanced.

So the balanced equation is: $\ce{1 Co(NO_{3})_{3} + 3 KOH -> 1 Co(OH)_{3} + 3 KNO_{3}}$

Answer:

The coefficients are 1 (for $\ce{Co(NO_{3})_{3}}$), 3 (for $\ce{KOH}$), 1 (for $\ce{Co(OH)_{3}}$), and 3 (for $\ce{KNO_{3}}$). So filling in the boxes: 1, 3, 1, 3.