# Question #a511c

Jan 12, 2018

$14.6$ grams

#### Explanation:

The solution of Iron (III) Chloride is 0.15 M,

So it contains $0.15$ moles of $F e C {l}_{3}$ in $1$ Litre of solution.

$\Rightarrow$ It Contains $\frac{0.15 \cdot 15}{1000} = 0.00225$ moles in 15 mL.

The solution of Sodium Carbonate is also 0.15 M.

So, it contains $\frac{0.15 \cdot 20}{1000} = 0.003$ moles of $N a O H$ in 20 mL.

Lets Check The Undergoing Equation Now.

$3 N {a}_{2} C {O}_{3} + 2 F e C {l}_{3} = F {e}_{2} {\left(C {O}_{3}\right)}_{3} + 6 N a C l$

So, 3 moles of $N {a}_{2} C {O}_{3}$ react with 2 moles of $F e C {l}_{3}$ to form 1 mole of $F {e}_{2} {\left(C {O}_{3}\right)}_{3}$.

So, The liniting reactant is indeed the sodium carbonate.

So, If The Sodium Carbonate Runs Out, The Reaction will stop.

Let's Approach with the simplest method in existence, The One and Only, Unitary Method.

$3$ moles of $N {a}_{2} C {O}_{3} \rightarrow 1$ mole of $F {e}_{2} {\left(C {O}_{3}\right)}_{3}$

$1$ mole of $N {a}_{2} C {O}_{3} \rightarrow \frac{1}{3}$ moles of $F {e}_{2} {\left(C {O}_{3}\right)}_{3}$

$0.15$ moles of $N {a}_{2} C {O}_{3} \rightarrow \frac{0.15}{3}$ mole of $F {e}_{2} {\left(C {O}_{3}\right)}_{3}$ = $0.05$ mole of $F {e}_{2} {\left(C {O}_{3}\right)}_{3}$

= $\left(0.05 \cdot \left(56 \cdot 2 + 3 \cdot 60\right)\right)$ g of $F {e}_{2} {\left(C {O}_{3}\right)}_{3}$

= $14.6$ g of $F {e}_{2} {\left(C {O}_{3}\right)}_{3}$

Jan 12, 2018

$\text{0.29 grams}$.

#### Explanation:

First, start off with the skeleton equation:
$F e C {l}_{3} + N {a}_{2} C {O}_{3} \to F {e}_{2} {\left(C {O}_{3}\right)}_{3} + N a C l$

Then, balance the equation to determine the mole ratios:
$2 F e C {l}_{3} + 3 N {a}_{2} C {O}_{3} \to F {e}_{2} {\left(C {O}_{3}\right)}_{3} + 6 N a C l$

We now know that, in this reaction, for every $2$ moles of iron(III) chloride, there will be $3$ moles of sodium carbonate, $1$ mole of iron(III) carbonate, and $6$ moles of sodium chloride.

If we know the number of moles for one reactant, we can predict the number of moles for all the other reactants and products.
However, in this problem, there's a twist—amounts for two reactants are given.
This means that one of these reactants is a limiting reactant.

To find out which is the limiting reactant, first determine the number of moles that react for each reactant.

1. For iron(III) chloride, $\text{0.00225 moles}$.
$\text{0.15 moles = 1000 mL}$
$\text{15 mL"= 0.15*15/1000 = "0.00225 moles}$.
2. For sodium carbonate, $\text{0.003 moles}$.
$\text{1000 mL = 0.15 moles}$
$\text{20 mL" = 0.15 * 20/1000 = "0.003 moles}$

Recall that in the mole ratio, $2$ moles of iron(III) chloride will react for every $3$ moles of sodium carbonate.

To react with $\text{0.003 moles}$ of sodium carbonate, $0.003 \cdot \frac{2}{3} = \text{0.002 moles}$ of iron(III) chloride is needed.
We have $\text{0.00225 moles}$. This means iron(III) chloride is not the limiting reactant.

To react with $\text{0.00225 moles}$ of iron(III) chloride, $0.00225 \cdot \frac{3}{2} = \text{0.003375 moles}$ of sodium carbonate is needed.
We only have $\text{0.003 moles}$. This means sodium carbonate is the limiting reactant.

Then, just use the moles of the limiting reactant to find out the moles of the product. We use the limiting reactant because all of it reacts.

Every $3$ moles of sodium carbonate corresponds to $1$ mole of iron(III) carbonate. Therefore, $\text{0.003 moles}$ will correspond to $\text{0.001 moles}$ of iron(III) carbonate.

The molar mass of iron(III) carbonate is $2 \cdot 55.85 + 3 \cdot \left(12.01 + 3 \cdot 16.00\right) = \text{291.73 grams per mole}$
$\text{0.001 moles}$ will be $291.73 \cdot 0.001 = \text{0.29 grams}$.