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3 years ago

What is the limiting reactant if 8 g of Ba reacts with 2.8 g of Al2(SO4)3?

Chemistry

1 answer:

Nata [24]3 years ago

5 0

Answer:

Al2(SO4)3 is the limiting reactant

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

3Ba + Al2(SO4)3 → 2Al + 3BaSO4

Next, we shall determine the mass of Ba and the mass of Al2(SO4)3 that reacted from the balanced equation. This is illustrated below:

Molar mass of Ba = 137g/mol

Mass of Ba from the balanced equation = 3 x 137 = 411g

Molar mass of Al2(SO4)3 = 2x27 + 3[32 + (16x4)]

= 54 + 3[32 + 64]

= 54 + 3[96]

= 54 + 288 = 342g/mol

Mass of Al2(SO4)3 from the balanced equation = 1 x 342 = 342g

Summary:

From the balanced equation above,

411g of Ba reacted with 342g of Al2(SO4)3.

Finally, we shall determine the limiting reactant as follow:

From the balanced equation above,

411g of Ba reacted with 342g of Al2(SO4)3.

Therefore, 8g of Ba will react with

= (8 x 342/411 = 6.66g of Al2(SO4)3.

From the calculations made above, we can see that it will take a higher mass of Al2(SO4)3 i.e 6.66g than what was given i.e 2.8g to react completely with 8g of Ba.

Therefore, Al2(SO4)3 is the limiting reactant and Ba is the excess reactant.

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Answer:

Limiting reactant = B2O3

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Explanation:

The given reaction is:

$B2O3 (s) + 3C (s) + 3Cl2 (g) \rightarrow 2BCl3 (g) + 3CO (g)$

In order to identify the limiting reagent calculate the moles of B2O3, C and Cl2. The reagent with the lowest moles is the limiting reactant

$Moles(B2O3)=\frac{Mass(B2O3)}{Mol.wt(B2O3)}=\frac{139g}{69.6g/mol}=1.997moles$

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Since the moles of B2O3 < C < Cl2, the limiting reactant is B2O3

Based on the reaction stoichiometry:

1 mole of B2O3 produces 2 moles of BCl3

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3 years ago

What is the molarity of a solution containing 6.0 moles of solute in 549 mL of solution?

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How many molecules of co2 in 97.3 grams of co2

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Answer:

1.33 × 10²⁴ molecules CO₂

General Formulas and Concepts:

Chemistry - Stoichiometry

Reading a Periodic Table
Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

Step 1: Define

97.3 g CO₂

Step 2: Define conversions

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Molar Mass of C - 12.01 g/mol

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Molar Mass of CO₂ - 12.01 + 2(16.00) = 44.01 g/mol

Step 3: Convert

$97.3 \ g \ CO_2(\frac{1 \ mol \ CO_2}{44.01 \ g \ CO_2} )(\frac{6.022 \cdot 10^{23} \ molecules \ CO_2}{1 \ mol \ CO_2} )$ = 1.33138 × 10²⁴ molecules CO₂

Step 4: Check

We are given 3 sig figs. Follow sig fig rules.

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