Question

1. balance the double replacement equation by typing the numbers in the blanks. type in 1 if the coefficient is 1.

pb(no₃)₂ + k₂s → pbs + kno₃

1. balance the double replacement equation by typing the numbers in the blanks. type in 1 if the coefficient is 1.

fecl₃ + naoh → fe(oh)₃ + nacl
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Explanation:

Question 7:

Step1: Balance Pb

On the left, we have 1 Pb in $\ce{Pb(NO_{3})_{2}}$, and on the right, 1 Pb in $\ce{PbS}$. So Pb is balanced initially (coefficient 1 for both).

Step2: Balance S

Left: 1 S in $\ce{K_{2}S}$, right: 1 S in $\ce{PbS}$. So S is balanced (coefficient 1 for $\ce{K_{2}S}$ and $\ce{PbS}$).

Step3: Balance K and $\ce{NO_{3}^-}$

Left: 2 K in $\ce{K_{2}S}$, right: 1 K in $\ce{KNO_{3}}$. So we need 2 $\ce{KNO_{3}}$ to balance K. Then, check $\ce{NO_{3}^-}$: left has 2 $\ce{NO_{3}^-}$ in $\ce{Pb(NO_{3})_{2}}$, right has 2 $\ce{NO_{3}^-}$ in 2 $\ce{KNO_{3}}$. So coefficients: 1, 1, 1, 2.

Step1: Balance Fe

Left: 1 Fe in $\ce{FeCl_{3}}$, right: 1 Fe in $\ce{Fe(OH)_{3}}$. Fe is balanced (coefficient 1 for both).

Step2: Balance Cl

Left: 3 Cl in $\ce{FeCl_{3}}$, right: 1 Cl in $\ce{NaCl}$. So we need 3 $\ce{NaCl}$ to balance Cl.

Step3: Balance Na and $\ce{OH^-}$

Left: 1 Na in $\ce{NaOH}$, right: 3 Na in 3 $\ce{NaCl}$. So we need 3 $\ce{NaOH}$ to balance Na. Then, check $\ce{OH^-}$: left has 3 $\ce{OH^-}$ in 3 $\ce{NaOH}$, right has 3 $\ce{OH^-}$ in $\ce{Fe(OH)_{3}}$. So coefficients: 1, 3, 1, 3.

Answer:

1, 1, 1, 2

Question 8: