A solution containing 40 g of calcium bromide requires 14.2 g of chlorine to react completely with it and 22,2 g of calcium chloride is produced in addition to ehaever bromine is obtained. How many grams of calcium bromide are required to produce 10.0 pounds of bromine?

convert the mass to moles

write the balanced equation
see how much excess Br remains from 40g of CaBr2 -- call that x grams

Multiply that by 10 lbs/x g

Good practice in unit conversions!

First, we need to convert pounds to grams. Since 1 pound is equal to 453.6 grams, we can calculate:

10.0 pounds * 453.6 grams/pound = 4536 grams

To find the molar mass of calcium bromide (CaBr2), we add the atomic masses of calcium and bromine:

1 calcium atom = 40.08 g/mol
2 bromine atoms = 2 * 79.90 g/mol = 159.8 g/mol
Total molar mass of CaBr2 = 40.08 g/mol + 159.8 g/mol = 199.88 g/mol

Now, we can set up the following proportion:

40 g of CaBr2 / 14.2 g of Cl2 = x grams of CaBr2 / 4536 g of Br2

Cross multiplying, we get:

(40 g of CaBr2) * (4536 g of Br2) = (14.2 g of Cl2) * (x grams of CaBr2)

Simplifying, we have:

181440 g of Br2 = 14.2 g of Cl2 * x grams of CaBr2

Dividing both sides of the equation by 14.2 g of Cl2, we get:

x grams of CaBr2 = (181440 g of Br2) / (14.2 g of Cl2) = 12796.33 g

Therefore, around 12796.3 grams of calcium bromide are required to produce 10.0 pounds of bromine.

To find out how many grams of calcium bromide are required to produce 10.0 pounds of bromine, we can follow these steps:

Step 1: Calculate the molar mass of calcium bromide (CaBr2).
The molar mass of calcium (Ca) is 40.08 g/mol, and the molar mass of bromine (Br) is 79.90 g/mol. Since calcium bromide (CaBr2) contains two bromine atoms, we can calculate the molar mass as follows:
Molar mass of CaBr2 = (Molar mass of Ca) + 2 * (Molar mass of Br)
= 40.08 g/mol + 2 * 79.90 g/mol
≈ 199.88 g/mol

Step 2: Calculate the amount of calcium bromide required to produce 10.0 pounds of bromine.
To convert pounds to grams, we need to know the conversion factor between pounds and grams. The conversion factor is 1 pound = 453.592 grams.

10.0 pounds * 453.592 grams/pound = 4535.92 grams

Step 3: Use the stoichiometry of the balanced chemical equation to determine the amount of calcium bromide required.
Based on the given information, we know that 40 g of calcium bromide reacts with 14.2 g of chlorine to produce 22.2 g of calcium chloride and bromine.

From the balanced chemical equation:
2 CaBr2 + 3 Cl2 → 2 CaCl2 + 2 Br2

We can set up a proportion to solve for the unknown amount of calcium bromide:
(40 g calcium bromide / 14.2 g chlorine) = (x g calcium bromide / 22.2 g bromine)

Simplifying the proportion:
(40 / 14.2) = (x / 22.2)

Solving for x:
x = (40 / 14.2) * 22.2
x ≈ 62.39 g

Therefore, approximately 62.39 grams of calcium bromide are required to produce 10.0 pounds of bromine.