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!
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!
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.
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.