Question

Use the balanced equation to solve the problem.Fe(NO3)3 + 3NaOH Fe(OH)3 + 3NaNO31196g Fe(OH)3 are made in this precipitation reaction.How many grams of NaOH reacted?g

Answer 1

The mass of NaOH that reacted is 219.58 grams

Step-by-step explanation

Given that

The mass of Fe(OH)3 is 196 grams

Follow the steps below to find the mass of NaOH

Step 1; Write the balanced equation of the reaction

`\text{ Fe\lparen NO}_3)_3\text{ + 3NaOH }\rightarrow\text{ Fe\lparen OH\rparen}_3\text{ + 3NaNO}_3`

Step 2; Find the moles of Fe(OH)3 using the below formula

`\text{ mole = }\frac{\text{ mass}}{\text{ molar mass}}`

Recall, that the molar mass of Fe(OH)3 is 106.87 g/mol

`\begin{gathered} \text{ mole = }\frac{\text{ 196}}{\text{ 106.87}} \\ \text{ mole = 1.83 moles} \end{gathered}`

The number of moles of Fe(OH)3 is 1.83 moles

Step 3; Find the number of moles of NaOH using a stoichiometry ratio

In the equation above, 1 mole of Fe(OH)3 gives 3 moles of NaOH

Let x represents the number of moles of Fe(OH)3

`\begin{gathered} \text{ 1 mole Fe\lparen OH\rparen}_3\text{ }\rightarrow\text{ 3 moles NaOH} \\ \text{ 1.83 moles Fe\lparen OH\rparen}_3\text{ }\rightarrow\text{ x moles NaOH} \\ \text{ Cross multiply} \\ \text{ 1 mole Fe\lparen OH\rparen}_3\text{ }*\text{ x moles NaOH = 1.83 moles Fe\lparen OH\rparen}_3*3\text{ moles NaOH} \\ \text{ Isolate x moles NaOH} \\ \text{ x moles NaOH = }\frac{1.83moles\cancel{Fe(OH)\placeholder{⬚}_3}*\text{ 3 moles NaOH}}{1mole\cancel{Fe(OH)\placeholder{⬚}_3}} \\ \text{ x moles NaOH = 1.83 }*\text{ 3} \\ \text{ x moles NaOH = 5.49 moles} \end{gathered}`

The number of moles of NaOH is 5.49 moles

Step 4; Find the mass of NaOH using the below formula

`\begin{gathered} \text{ mole = }\frac{\text{ mass}}{\text{ molar mass}} \\ \text{ cross multiply} \\ \text{ mass = mole }*\text{ molar mass} \end{gathered}`

Recall, that the molar mass of NaOH is 39.997 g/mol

`\begin{gathered} \text{ mass = 5.49 }*\text{ 39.997} \\ \text{ mass = 219.58 grams} \end{gathered}`

Therefore, the mass of NaOH that reacted is 219.58 grams