A sample of 4.50 g of methane is mixed with 14.0 g of chlorine . Determine which

Question

A sample of 4.50 g of methane  is mixed with 14.0 g of chlorine .

Determine which is the limiting reactant according to the following equation:

The limiting reactant is

.

What is the maximum mass of  that can be formed?

Pad? 3:56 PM * 13%D+ sjc.cengagenow.com Wendy Avila owLv2 l Online teachi ing and learn.Chegg Study Guided Solutions and.. extra credit" Ch 5 EOC A sample of 4.50 g of methane (CH4) is mixed with 14.0 g of chlorine (C2) Question 2 1 pt Question 3 a. Determine which is the limiting reactant according to the following equation Question 4 1 pt CH4 (g) + 4Cl2 (g)CC14 (e) + 4HCl(g) Question 5 b. The limiting reactant is Question 6 1 pt What is the maximum mass of CC14 that can be formed? Question 7 Maximum mass Question 8 1 pt Question 9 Submit Answer Try Another Version 1 pt Question 10 Question 11 Question 12 Question 13 Question 14 Question 15 Question 16 Question 17 Question 18 Question 19 Question 20 5 item attempts remaining 1 pt 1 pt 1 pt 1 pt Progress 10/20 items Due Apr 1 at 11:55 PM Previous Next Finish Assignment Save and Exit Cengage Learning| Cengage Technical Support

Explanation / Answer

Number of moles of CH4 = 4.50 g / 16.04 g/mol = 0.281 mole

number of moles of chlorine = 14.0 g / 70.9060 g/mol = 0.197 mole

from the balanced equation we can say that

1 mole of CH4 requires 4 mole of Cl2 so

0.281 mole of CH4 will require

= 0.281 mole of CH4 *(4 mole of Cl2 /1 mole of CH4 )

= 1.12 mole of Cl2

But we have 0.197 mole of Cl2 only which is in less amoune so chlorine is the limiting reactant

The limiting reactant is Cl2

From the balanced equation we can say that

4 mole of Cl2 produces 1 mole of CCl4 so

1.12 mole of Cl2 will produce 0.28 mole of CCl4

mass of 1 mole of CCl4 = 153.82 g

so the mass of 0.280 mole of CCl4 = 43.1 g

Therefore, the mass of CCl4 produced would be 43.1 g

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