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For the following reaction, 25.4 grams of carbon disulfide are allowed to react with 98.9 grams of chlorine gas.
carbon disulfide (s) + chlorine (g) carbon tetrachloride (l) + sulfur dichloride (s)
1. What is the maximum amount of carbon tetrachloride that can be formed? _________grams
2. What is the FORMULA for the limiting reagent?_________
3. What amount of the excess reagent remains after the reaction is complete? ________grams

Respuesta :

Answer:

1) 51.37 grams CCl4

2) CS2

3) There remains 0.393 moles = 27.87 grams Cl2

Explanation:

Step 1: Data given

Mass of carbon disulfide = 25.4 grams

Mass of chlorine gas = 98.9 grams

Molar mass of carbon disulfide = 76.14 g/mol

Molar mass of chlorine gas = 70.91 g/mol

Molar mass of Carbon tetrachloride = 153.81 g/mol

Step 2: The balanced equation:

CS2 + 3Cl2 → CCl4 + S2Cl2

For 1 mol CS2 consumed, we need 3 moles Cl2 to produce 1 mole CCl4 and 1 mole S2Cl2

Step 3: Calculate moles CS2

Number of moles CS2= mass CS2 / Molar mass CS2

Moles CS2 = 25.4 grams / 76.14 g/mol

Moles CS2 =  0.334 moles

Step 4: Calculate moles Cl2

Moles Cl2 = 98.9 grams / 70.91 g/mol

Moles Cl2 = 1.395 moles

Step 5: The limiting reactant

CS2 is the limiting reactant, it will completely consumed ( 0.334 moles).

Cl2 is in excess. There will be consumed 3*0.334 = 1.002 moles

There will remain 1.395 - 1.002 = 0.393 moles Cl2

0.393 moles Cl2 = 27.87 grams Cl2

Step 6: Calculate moles CCl4

For 1 mol CS2 consumed, we need 3 moles Cl2 to produce 1 mole CCl4 and 1 mole S2Cl2

For 0.334 moles CS2 we have 0.334 moles CCl4

Step 7: Calculate mass of CCl4

Mass CCl4 = Moles CCl4 * Molar mass CCl4

Mass CCl4 = 0.334 moles * 153.81 g/mol

Mass CCl4 = 51.37 grams

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