Stoikiometri Pereaksi Pembatas 1

Summary

The video explains how to calculate the number of moles of NaCl formed from a reaction between NaOH and HCl. It begins by discussing the balanced chemical reaction and introduces the concept of limiting reagents. The video guides through step-by-step calculations of moles for NaOH and HCl, identifies the limiting reagent, and calculates the moles of NaCl produced. It also covers a second problem involving ammonia production, balancing the reaction, identifying the limiting reagent, and calculating the amount of NH3 formed and the leftover hydrogen.

Takeaways

• 🧪 The reaction equation given is NaOH + HCl → NaCl + H2O, which is already balanced with a 1:1 ratio of reactants.
• 🔢 To find the moles of NaOH, multiply the concentration (1.5 M) by the volume (100 mL), resulting in 15 mmol of NaOH.
• ⚗️ Similarly, to find the moles of HCl, multiply the concentration (0.1 M) by the volume (100 mL), resulting in 10 mmol of HCl.
• 📉 The limiting reactant is determined by dividing the initial moles by the stoichiometric coefficients. HCl is the limiting reactant because 10 mmol is the smallest value.
• 🧮 10 mmol of HCl reacts completely with 10 mmol of NaOH, leaving 5 mmol of NaOH unreacted.
• 🧂 The reaction produces 10 mmol of NaCl as per the 1:1 molar ratio between HCl and NaCl.
• 💧 The reaction also produces 10 mmol of water, although this can be ignored for most calculations.
• 🌡️ For a second reaction (N2 + H2 → NH3), the equation is balanced as N2 + 3H2 → 2NH3.
• ⚖️ In this case, 0.5 mol of N2 and 2.4 mol of H2 are given, and N2 is the limiting reactant since 0.5 mol is less than 0.8 mol when divided by their coefficients.
• 💥 The limiting reactant (N2) reacts completely, producing 1 mol of NH3, and leaves 0.9 mol of H2 unreacted.

Q & A

What is the chemical reaction described in the transcript?

The chemical reaction described is NaOH + HCl → NaCl + H2O, which is a neutralization reaction between sodium hydroxide and hydrochloric acid, resulting in sodium chloride (NaCl) and water (H2O).

How is the mol of NaOH calculated from the given concentration and volume?

The mol of NaOH is calculated by multiplying its concentration (1.5 M) with the volume (100 mL), giving 1.5 × 0.1 = 0.15 mol.

What is the limiting reagent in the NaOH and HCl reaction?

The limiting reagent is HCl because its initial mol (10 mmol) is less than that of NaOH (15 mmol), and thus it is completely consumed first.

How many mols of NaCl are produced in the reaction?

10 mmol of NaCl are produced, as the reaction between NaOH and HCl has a 1:1 molar ratio, so the amount of NaCl formed is equal to the limiting reagent (HCl), which is 10 mmol.

What is the final amount of NaOH left after the reaction?

After the reaction, 5 mmol of NaOH remain unreacted. This is calculated by subtracting the amount of NaOH that reacted (10 mmol) from the initial 15 mmol.

What is the role of the reaction coefficients in determining how much of each reactant is used?

The reaction coefficients help in determining the mole ratio of reactants and products. In this case, the reaction between NaOH and HCl has a 1:1 ratio, so equal mols of NaOH and HCl react with each other.

In the second reaction, what is the balanced equation for the production of ammonia (NH3)?

The balanced equation for the production of ammonia is N2 + 3H2 → 2NH3.

How is the limiting reagent determined in the reaction for NH3 production?

The limiting reagent is determined by dividing the initial mol of each reactant by its coefficient. For N2, it's 0.5 mol / 1 = 0.5, and for H2, it's 2.4 mol / 3 = 0.8. Since 0.5 is smaller, N2 is the limiting reagent.

How many mols of NH3 are produced in the ammonia reaction?

1 mol of NH3 is produced. Since 0.5 mol of N2 reacts and the mole ratio between N2 and NH3 is 1:2, the amount of NH3 formed is twice the amount of N2, which is 1 mol.

What is the amount of H2 left unreacted in the NH3 production reaction?

After the reaction, 0.9 mol of H2 remain unreacted. This is calculated by subtracting the amount that reacted (1.5 mol) from the initial 2.4 mol.