Aluminum metal reacts with chlorine with a spectacular display of sparks: 2Al(s)+3Cl2(g)→2AlCl3(s) ΔH∘ = -1408.4 kJ....

Question

Question

Aluminum metal reacts with chlorine with a spectacular display of sparks: 2Al(s)+3Cl2(g)→2AlCl3(s) ΔH∘ = -1408.4 kJ.

How much heat (in kilojoules) is released on reaction of 5.55 g of Al?

science chemistry

Answer 1

Answer #1

Answer 2

Similar Homework Help Questions

Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl2(g)?2AlCl3(s) What is...

Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl2(g)?2AlCl3(s) What is the maximum mass of aluminum chloride that can be formed when reacting 27.0 g of aluminum with 32.0 g of chlorine?
Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl2(g)?2AlCl3(s) What is...

Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl2(g)?2AlCl3(s) What is the maximum mass of aluminum chloride that can be formed when reacting 17.0g of aluminum with 22.0g of chlorine?
Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl2(g)→2AlCl3(s) You are...

Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl2(g)→2AlCl3(s) You are given 16.0 g of aluminum and 21.0 g of chlorine gas. If you had excess aluminum, how many moles of aluminum chloride could be produced from 21.0 g of chlorine gas, Cl2? Express your answer to three significant figures and include the appropriate units.

Aluminum reacts with chlorine gas to form aluminum chloride. 2Al(s)+3Cl2(g)→2AlCl3(s) What minimum volume of chlorine gas...

Aluminum reacts with chlorine gas to form aluminum chloride. 2Al(s)+3Cl2(g)→2AlCl3(s) What minimum volume of chlorine gas (at 298 KK and 217 mmHgmmHg) is required to completely react with 8.48 gg of aluminum?
he balanced equation for the reaction of aluminum metal and chlorine gas is 2Al(s) + 3Cl2(g)...

he balanced equation for the reaction of aluminum metal and chlorine gas is 2Al(s) + 3Cl2(g) → 2AlCl3(s) Assume that 0.43 g Al is mixed with 0.54 g Cl2. (a) What is the limiting reactant? Cl2 or Al (b) What is the maximum amount of AlCl3, in grams, that can be produced?
Aluminum reacts with chlorine gas to form aluminum chloride. 2Al(s)+3Cl2(g)→2AlCl3(s) What minimum volume of chlorine gas...

Aluminum reacts with chlorine gas to form aluminum chloride. 2Al(s)+3Cl2(g)→2AlCl3(s) What minimum volume of chlorine gas (at 298 K and 212 mmHg ) is required to completely react with 7.94 g of aluminum? What minimum volume of chlorine gas (at 298 and 212 ) is required to completely react with 7.94 of aluminum? 387 L 0.387 L 38.7 mL 38.7 L

Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl2(g)→2AlCl3(s) You are...

Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl2(g)→2AlCl3(s) You are given 13.0 g of aluminum and 18.0 g of chlorine gas. Part A If you had excess chlorine, how many moles of of aluminum chloride could be produced from 13.0 g of aluminum? Express your answer to three significant figures and include the appropriate units. If you had excess aluminum, how many moles of aluminum chloride could be produced from 18.0 g of chlorine...
2.02 g of aluminum metal reacts with 1.71L of chlorine gas at 300K and 1.20 atm...

2.02 g of aluminum metal reacts with 1.71L of chlorine gas at 300K and 1.20 atm to produce 3.51 g of aluminum chloride. Determine the precent yield for this reaction. 2Al (s) + 3Cl2(g) -----> 2AlCl3 (aq)
Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s) + 3Cl2(g) +2AICI3(s)

Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s) + 3Cl2(g) +2AICI3(s) You are given 12.0 g of aluminum and 170 g of chlorine gas. Part A If you had excess chlorine, how many moles of aluminum chloride could be produced from 12.0 g of aluminum?

a. 2Al(s)+3Cl2(g)→2AlCl3(s) You are given 19.0 g of aluminum and 24.0 g of chlorine gas.If you had excess aluminum, how...

a. 2Al(s)+3Cl2(g)→2AlCl3(s) You are given 19.0 g of aluminum and 24.0 g of chlorine gas.If you had excess aluminum, how many moles of aluminum chloride could be produced from 24.0 g of chlorine gas, Cl2? b. Determine the balanced chemical equation for this reaction. C8H18(g)+O2(g)→CO2(g)+H2O(g) c. 3H2(g)+N2(g)→2NH3(g) 1.08 g H2 is allowed to react with 10.3 g N2, producing 1.04 g NH3. What is the theoretical yield for this reaction under the given conditions? What is the percent yield for...