Question

If \(120 \mathrm{gm}\) of propanol reacts with \(23 \mathrm{gm}\) of \(\mathrm{Na}\), then what will be the weight of released gas? (A) \(2 \mathrm{gm}\) (B) \(60 \mathrm{gm}\) (C) \(1 \mathrm{gm}\) (D) \(1.5 \mathrm{gm}\)

Short answer

The weight of the released gas will be approximately \(1 \ \mathrm{gm}\). The answer is (C) \(1 \mathrm{gm}\).

Step-by-step solution

Write down the balanced chemical equation for the reaction

First, we need the method to write down the balanced chemical equation for the reaction between propanol (C3H7OH) and sodium (Na). Propanol is an alcohol and sodium is a metal, the general reaction between an alcohol and a metal can be represented by: \( 2\ \mathrm{C_3H_7OH + 2\ Na \rightarrow 2\ C_3H_7ONa + H_2} \) This shows that for every 2 moles of propanol reacting with 2 moles of sodium, one mole of hydrogen gas is released.

Calculate the moles of propanol and sodium

We need to calculate the moles of propanol and sodium given their weights in the problem statement. Moles of propanol = \( \frac{Weight\ of\ Propanol}{Molecular\ weight\ of\ Propanol} \) Molecular weight of propanol (C3H7OH) = 3(12.01)+7(1.01)+15.99+1.01 = 60.11 g/mol Moles of propanol = \( \frac{120 \ \mathrm{gm} }{60.11 \ \mathrm{gm\ / \ mol} } = 1.998 \ \mathrm{moles} \) Moles of sodium = \( \frac{Weight \ of\ Sodium}{Molecular\ weight\ of\ Sodium} \) Molecular weight of sodium (Na) = 23.00 g/mol Moles of sodium = \( \frac{23 \ \mathrm{gm}}{23.00 \ \mathrm{gm\ / \ mol}} = 1 \ \mathrm{mole} \)

Determine the limiting reactant

Next, we need to determine which reactant is the limiting reactant. The balanced chemical equation shows that 2 moles of propanol react with 2 moles of sodium. Therefore, for the reaction to go to completion, we need equal mole ratios of propanol and sodium. Since we have 1.998 moles of propanol and 1 mole of sodium, the reaction will be limited by the amount of sodium. Hence, sodium is the limiting reactant.

Calculate the released gas weight

Use the balanced chemical equation to find the moles of released hydrogen gas. From the equation, we can see that 2 moles of sodium produce 1 mole of hydrogen gas: \( 1 \ \mathrm{mole\ of\ Na} \times \frac{1 \ \mathrm{mole\ H_2}}{2 \ \mathrm{moles\ of\ Na}} = 0.5 \ \mathrm{moles\ of\ H_2} \) Now, convert the moles of hydrogen gas into grams: Weight of released hydrogen gas = Moles of hydrogen gas × Molecular weight of hydrogen gas = 0.5 moles × (2 × 1.01 gm/mol) = 1.01 gm The weight of released gas is approximately 1 gm. The answer is (C) \(1 \mathrm{gm}\).

Key concepts

Limiting reactant

In any chemical reaction, the limiting reactant is the substance that gets completely consumed first, halting the reaction as no further products can be formed. Identifying the limiting reactant is crucial as it determines the amount of product that can be generated.
In this exercise, we start by looking at the balanced equation of the reaction between propanol and sodium:

• For every 2 moles of propanol (\( \text{C}_3\text{H}_7\text{OH} \)) reacting with 2 moles of sodium (\( \text{Na} \)), 1 mole of hydrogen gas (\( \text{H}_2 \)) is released.

When calculating moles, we find there are about 1.998 moles of propanol and exactly 1 mole of sodium present. Comparing this ratio to the balanced equation, sodium becomes the limiting reactant as it reaches its required proportion (2 moles for 2 moles of propanol) first.
This means the reaction will proceed until all sodium is used up. Once sodium is consumed, no further products can be generated. Understanding this concept helps in predicting the actual yield of the reaction.

Molar mass calculation

Molar mass is a critical component in understanding how much of a substance reacts or is produced in a chemical reaction. It refers to the mass of one mole of a substance and is often expressed in grams per mole (g/mol). Each element within the compound contributes to the overall molar mass based on its atomic weight and the number of atoms present.
In the exercise, the molar mass of propanol is needed. Propanol (\( \text{C}_3\text{H}_7\text{OH} \)) is comprised of:

• 3 Carbon (\( \text{C} \)) atoms each weighing \(12.01 \text{ g/mol}\),
• 7 Hydrogen (\( \text{H} \)) atoms each weighing \(1.01 \text{ g/mol}\),
• 1 Oxygen (\( \text{O} \)) atom weighing \(15.99 \text{ g/mol}\),
• 1 additional Hydrogen (\( \text{H} \)) atom for the \(-\text{OH}\) group weighing \(1.01 \text{ g/mol}\).

Adding these gives a molar mass of \(60.11 \text{ g/mol}\) for propanol. This calculation allows us to determine how much propanol reacts with sodium by converting grams to moles, enabling us to use these figures in stoichiometry for predicting product amounts.

Balanced chemical equation

A balanced chemical equation is a fundamental aspect of understanding chemical reactions. It ensures the law of conservation of mass is applied, where mass is neither created nor destroyed in a reaction. Balancing equations involves making sure there are equal numbers of each type of atom on both sides of the equation.
Let's look at the equation for propanol and sodium:

• The reaction is \( 2\ \text{C}_3\text{H}_7\text{OH} + 2 \ \text{Na} \rightarrow 2\ \text{C}_3\text{H}_7\text{ONa} + \text{H}_2 \).
• This equation is balanced because:
  • There are 6 carbons, 16 hydrogens, 2 oxygens, and 2 sodiums on both sides of the reaction.
• Balancing the equation correctly reflects the fixed ratios in which reactants turn into products.

Balanced chemical equations are crucial for computing reactant consumption and product formation accurately, forming the basis of stoichiometry calculations. Properly writing and balancing equations helps predict how much reactant is needed and how much product is expected, streamlining chemical analysis processes.