Mole Concept

Question

The limiting reagent in the reaction 2Al + 3Cl2 → 2AlCl3 when 10.0 g Al (M=27) reacts with 20.0 g Cl2 (M=71) is:

RankGuru · Accepted Answer

1. **Balanced Equation**: $2 \ Al + 3 \ Cl_2 	o 2 \ AlCl_3$.
2. **Calculate Initial Moles**:
   - $n_{Al} = \frac{10.0}{27} \approx 0.370 \ mol$.
   - $n_{Cl_2} = \frac{20.0}{71} \approx 0.282 \ mol$.
3. **Determine Stoichiometric Requirement**:
   - According to the reaction, $1 \ mol$ of $Al$ requires $1.5 \ mol$ of $Cl_2$
   - For $0.370 \ mol \ Al$, required $Cl_2 = 0.370 	imes \frac{3}{2} = 0.555 \ mol$.
4. **Identify Limiting Reagent**:
   - Since the required $Cl_2$ ($0.555 \ mol$) is greater than the available $Cl_2$ ($0.282 \ mol$),
   - **$Cl_2$ is the limiting reagent**.

Rank Guru

Mole Concept

Select the correct option:

Solution

🔒 Solution Hidden from View

More mole concept Practice Questions

How many moles of oxygen atoms are present in 5.0 g of CaCO3? (Atomic masses: Ca=40, C=12, O=16)

The density of a gas at STP is 1.25 g L⁻¹. Its molar mass is approximately:

When 9.0 g of aluminium reacts completely with oxygen, how many grams of Al2O3 are formed? (2Al + 3/...

Which statement correctly defines molality?

0.50 mol of CaCO3 is heated strongly. How many liters of CO2 (at STP) are evolved? (CaCO3 → CaO + CO...

The mass percent of nitrogen in NH4 NO3 (Atomic masses: N=14, H=1, O=16) is:

About This Question