Question

Remember that the stoichiometric coefficients provide information on the relative amounts of reactant needed and product produced for a given reaction. As an example, the balanced reaction equation for the combustion of methane, CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l) This equation can be interpreted as: - 1 molecule of methane reacts with 2 molecules of oxygen to produce 1 molecule of carbon dioxide and 2 molecules of water. - 1 mole of methane reacts with 2 moles of oxygen to produce 1 mole of carbon dioxide and 2 molecules of water. - 16 grams of methane react with 64 grams of oxygen to produce 44 grams of carbon dioxide and 36 grams of water. In this conference sheet, we will use stoichiometry to determine the limiting reactant. _ Al (s) + _ Cl2 (g) → _ AlCl3 (s) 2. Identify the limiting reactant for each of the following assuming we start with 10.0 g of each reactant. Note that you must balance the equations first! Moles of AlCl3 produced if all 10.0 g of Al is reacted: Moles of AlCl3 produced if all 10.0 g of Cl2 is reacted: Limiting reactant: ______________ 3. Calculate the mass of AlCl3 that would form.

          Remember that the stoichiometric coefficients provide information on the relative amounts of reactant needed and product produced for a given reaction. As an example, the balanced reaction equation for the combustion of methane,

CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l)

This equation can be interpreted as:
- 1 molecule of methane reacts with 2 molecules of oxygen to produce 1 molecule of carbon dioxide and 2 molecules of water.
- 1 mole of methane reacts with 2 moles of oxygen to produce 1 mole of carbon dioxide and 2 molecules of water.
- 16 grams of methane react with 64 grams of oxygen to produce 44 grams of carbon dioxide and 36 grams of water.

In this conference sheet, we will use stoichiometry to determine the limiting reactant.

___ Al (s) + ___ Cl2 (g) → ___ AlCl3 (s)

2. Identify the limiting reactant for each of the following assuming we start with 10.0 g of each reactant. Note that you must balance the equations first! Moles of AlCl3 produced if all 10.0 g of Al is reacted:
Moles of AlCl3 produced if all 10.0 g of Cl2 is reacted:
Limiting reactant: ________________

3. Calculate the mass of AlCl3 that would form.

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Added by Bruce S.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

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Solved by Expert Suman K

05/09/2022

Step 1

Step 1: Balance the chemical equation: \[2 \text{Al (s)} + 3 \text{Cl}_2 (g) \rightarrow 2 \text{AlCl}_3 (s)\] Show more…

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Remember that the stoichiometric coefficients provide information on the relative amounts of reactant needed and product produced for a given reaction. As an example, the balanced reaction equation for the combustion of methane, CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l) This equation can be interpreted as: - 1 molecule of methane reacts with 2 molecules of oxygen to produce 1 molecule of carbon dioxide and 2 molecules of water. - 1 mole of methane reacts with 2 moles of oxygen to produce 1 mole of carbon dioxide and 2 molecules of water. - 16 grams of methane react with 64 grams of oxygen to produce 44 grams of carbon dioxide and 36 grams of water. In this conference sheet, we will use stoichiometry to determine the limiting reactant. ___ Al (s) + ___ Cl2 (g) → ___ AlCl3 (s) 2. Identify the limiting reactant for each of the following assuming we start with 10.0 g of each reactant. Note that you must balance the equations first! Moles of AlCl3 produced if all 10.0 g of Al is reacted: Moles of AlCl3 produced if all 10.0 g of Cl2 is reacted: Limiting reactant: ________________ 3. Calculate the mass of AlCl3 that would form.

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Transcript

00:02 Hello everyone.

00:04 So the question says that here there is a reaction given where aluminum is reacting with chlorine giving rise to aluminum chloride and we have been given that each reactant amount is given to us which is 10 gram.

00:39 Aluminum is 10 gram.

00:40 Chlorine is 10 gram.

00:43 So we have to find out the limiting reactant here and also we have to find out the mass of aluminium chloride that would be obtained from 10 gram of aluminium and 10 gram of chlorine.

00:56 So in order to find out these things we first need to balance the reaction.

01:05 Aluminum plus chlorine giving rise to aluminum chloride so after balancing the reaction will be like 2 aluminum, 3 mol of chlorine, that gives 2 mol of aluminum chloride.

01:21 So we can see here that 2 mole of aluminum is reacting with 3 mol of chlorine or we can say 3 mol of chlorine is reacting with 2 moles of aluminum.

01:37 So we know that 1 mole of aluminum weighs 1 mole of aluminum weighs, 1 mole of aluminum weighs 27.

01:54 1 mole of aluminum weighs 27 gram and or we can say that 27 gram of aluminum is 1 mole so 1 gram would be 1 divided by 27 moles 10 gram of aluminum will be 1 divided by 27 multiplied by 10 that gives that becomes 0 .37 moles or of aluminum.

02:38 Now the moles of chlorine present here, 1 mole of chlorine weighs, 1 mole of chlorine weighs 71 grams, 71 grams.

03:02 Or we can say 71 grams of chlorine is equal to 1 mole.

03:08 1 gram would be 1 divided by 71, but we have 10 grams of chlorine...

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