Problem 90 - Chemistry and Chemical Reactivity

Question

Aqueous solutions of iron(II) chloride and sodium sulfide react to form iron(11)sulfide and sodium chloride. (a) Write the balanced equation for the reaction. (b) If you combine \(40 .\) g each of \(\mathrm{Na}_{2} \mathrm{S}\) and \(\mathrm{FeCl}_{2}\), what is the limiting reactant? (c) What mass of FeS is produced? (d) What mass of NasS or FeCl, remains after the reaction? (e) What mass of \(\mathrm{FeCl}_{2}\) is required to react completely with \(40 . \mathrm{g}\) of \(\mathrm{Na}_{2} \mathrm{S} ?\)

Step-by-Step Solution

Answer

(a) \(\text{FeCl}_2 + \text{Na}_2\text{S} \rightarrow \text{FeS} + 2\text{NaCl}\); (b) \(\text{FeCl}_2\) is the limiting reactant; (c) \(27.52 \, \text{g} \text{of} \, \text{FeS}\) is produced; (d) \(15.53 \, \text{g} \, \text{of} \, \text{Na}_2\text{S}\) remains; (e) \(65.41 \, \text{g} \, \text{of} \, \text{FeCl}_2\) is needed.

Step 1: Write the Balanced Equation

First, we identify the reactants and products of the reaction. Iron(II) chloride, \(\text{FeCl}_2\), reacts with sodium sulfide, \(\text{Na}_2\text{S}\), to form iron(II) sulfide, \(\text{FeS}\), and sodium chloride, \(\text{NaCl}\). The unbalanced equation is: \[ \text{FeCl}_2 + \text{Na}_2\text{S} \rightarrow \text{FeS} + \text{NaCl} \]To balance the equation:- Balance Fe: 1 Fe is each side.- Balance Cl: We need 2 NaCl for the 2 Cl in FeCl2.- Balance Na: We have 2 Na in Na2S and thus need 2 NaCl.The balanced equation is:\[ \text{FeCl}_2 + \text{Na}_2\text{S} \rightarrow \text{FeS} + 2\text{NaCl} \]

Step 2: Calculate Molar Masses

Calculate the molar masses of the compounds involved:- \(\text{Na}_2\text{S}\): \( 2(22.99) + 32.07 = 78.05 \, \text{g/mol} \)- \(\text{FeCl}_2\): \( 55.85 + 2(35.45) = 127.75 \, \text{g/mol} \)- \(\text{FeS}\): \( 55.85 + 32.07 = 87.92 \, \text{g/mol} \)

Step 3: Convert Mass to Moles

Convert the given masses of each reactant to moles:- Moles of \(\text{Na}_2\text{S}\): \( \frac{40.0}{78.05} = 0.512 \, \text{mol} \)- Moles of \(\text{FeCl}_2\): \( \frac{40.0}{127.75} = 0.313 \, \text{mol} \)

Step 4: Identify the Limiting Reactant

The balanced equation tells us that 1 mol of \(\text{Na}_2\text{S}\) reacts with 1 mol of \(\text{FeCl}_2\).- Based on Step 3, \(\text{FeCl}_2\) (0.313 mol) would run out before \(\text{Na}_2\text{S}\) (0.512 mol).Thus, \(\text{FeCl}_2\) is the limiting reactant.

Step 5: Calculate Mass of FeS Produced

Using the moles of the limiting reactant (\(\text{FeCl}_2\)), calculate the mass of \(\text{FeS}\) produced:- Moles of \(\text{FeS}\) = moles of \(\text{FeCl}_2\) = 0.313 mol- Mass of \(\text{FeS}\): \( 0.313 \, \text{mol} \times 87.92 \, \text{g/mol} = 27.52 \, \text{g} \)

Step 6: Determine Remaining Mass of Na2S

Since \(\text{FeCl}_2\) is the limiting reactant, some \(\text{Na}_2\text{S}\) will remain:- Moles used of \(\text{Na}_2\text{S}\) = moles of \(\text{FeCl}_2\) = 0.313 mol- Remaining moles of \(\text{Na}_2\text{S}\): \( 0.512 - 0.313 = 0.199 \, \text{mol} \)- Remaining mass of \(\text{Na}_2\text{S}\): \( 0.199 \, \text{mol} \times 78.05 \, \text{g/mol} = 15.53 \, \text{g} \)

Step 7: Calculate Mass of FeCl2 Needed for Full Reaction with 40 g Na2S

To fully react with 40 g of \(\text{Na}_2\text{S}\), calculate the moles of \(\text{Na}_2\text{S}\) and find equivalent moles of \(\text{FeCl}_2\):- Moles of \(\text{Na}_2\text{S}\) from 40 g: \( 0.512 \, \text{mol} \)- Mass equivalent of \(\text{FeCl}_2\): \(0.512 \, \text{mol} \times 127.75 \, \text{g/mol} = 65.41 \, \text{g} \)

Key Concepts

Stoichiometry

Stoichiometry is the mathematical heart of chemistry. It deals with the relationships between the quantities of reactants and products in chemical reactions. Understanding stoichiometry is crucial for calculating how much of each substance is needed or produced in a reaction.

In a chemical reaction, the coefficients in the balanced equation tell you the ratios between the molecules or moles of reactants and products. For example, in the equation: \[ \text{FeCl}_2 + \text{Na}_2\text{S} \rightarrow \text{FeS} + 2\text{NaCl} \], it tells us that:

• 1 mole of \( \text{FeCl}_2 \) reacts with 1 mole of \( \text{Na}_2\text{S} \).
• This produces 1 mole of \( \text{FeS} \) and 2 moles of \( \text{NaCl} \).

This information is vital in understanding how much of each reactant is needed to fully react and how much product is produced. Without stoichiometric calculations, we might end up using either too much or too little of a substance, leading to inefficient reactions.

Limiting Reactant

The limiting reactant is the substance that determines how far a chemical reaction can proceed. It's like the short leg in a three-legged race, controlling the pace of the race.

To identify the limiting reactant, compare the mole ratios of the substances. In the provided example, we converted the mass of \( \text{Na}_2\text{S} \) and \( \text{FeCl}_2 \) into moles:

• Moles of \( \text{Na}_2\text{S} = 0.512 \),
• Moles of \( \text{FeCl}_2 = 0.313 \).

According to the balanced equation, they react in a 1:1 ratio, which means we need equal moles of both to proceed the reaction completely.

Because \( \text{FeCl}_2 \) has fewer moles compared to \( \text{Na}_2\text{S} \), it will run out first, making it the limiting reactant. This concept is crucial because once the limiting reactant is consumed, the reaction stops, determining the maximum amount of product formed.

Balancing Chemical Equations

Balancing chemical equations is like making sure both sides of a seesaw are equal. It involves making sure the number of each type of atom on the reactant side is the same as on the product side.

Take this initial reaction equation: \[ \text{FeCl}_2 + \text{Na}_2\text{S} \rightarrow \text{FeS} + \text{NaCl} \]. Note that it is not yet balanced because the numbers of chlorine and sodium atoms do not match on both sides.

Here's how to balance it:

• First, balance the iron (Fe) atoms.
• Next, check the chlorine (Cl) atoms. There are 2 Cl in \( \text{FeCl}_2 \), so we need 2 \( \text{NaCl} \).
• Finally, balance the sodium (Na) atoms, which are now also balanced at 2 on each side after the chlorine adjustment.

This gives us the balanced equation: \[ \text{FeCl}_2 + \text{Na}_2\text{S} \rightarrow \text{FeS} + 2\text{NaCl} \]. Balancing equations is a fundamental skill in chemistry, ensuring the law of conservation of mass is adhered to.

Molar Mass Calculation

The molar mass of a substance is the mass of one mole of its particles, and it's a crucial concept for converting between mass and moles in chemical calculations. A mole is Avogadro's number worth of particles, which is \( 6.022 \times 10^{23} \).

Let's calculate some molar masses used in the chemical reaction above:

• \( \text{Na}_2\text{S} \): Add twice the atomic mass of sodium (Na), 22.99, plus sulfur (S), 32.07, totaling \( 78.05 \, \text{g/mol} \).
• \( \text{FeCl}_2 \): Combine the mass of iron (Fe), 55.85, with chlorine (Cl), 35.45, times 2 since there are two chlorine atoms, equaling \( 127.75 \, \text{g/mol} \).
• \( \text{FeS} \): Sum the iron (Fe), 55.85, and sulfur (S), 32.07, resulting in \( 87.92 \, \text{g/mol} \).

Knowing how to calculate molar mass helps in transforming a given mass to moles using the formula: \[ \text{Moles} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}} \]. This conversion is essential for performing stoichiometric calculations accurately.