if 4.1g of Cr is heateed with 9.3g of cl, what mass of crcl3 will be produce
Find the caps key on your keyboard and use it.
This is a limiting reagent (LR) problem.
Here is a procedure that will work all of your LR problems.
1. Write and balance the equatin.
Cr + 3Cl2 ==> 2CrCl3
2a. Convert grams Cr to mols. mols = grams/atomic mass = ?
2b. Do the same and convert grmas Cl2 to mols.
3a. Using the coefficients in the balanced equation, convert mols Cr to mols CrCl3.
3b. Do the same and convert mols Cl2 to mols CrCl3.
3c. It is likely that the values from 3a and 3b will not agree. In LR problems the correct value is ALWAYS the smaller one and the reagent responsible for that is called the LR.
4. Using the smaller value, convert mols to grams.
mols CrCl3 x molar mass CrCl3 = grams CrCl3.
Post your work if you get stuck.
i got 12.5g
am i right?
Right !. Good work. By the way, that equation I wrote is not balanced. It should be
2Cr + 3Cl2 ==>2CrCl3
Obviously you corrected that before you worked the problem.
To determine the mass of CrCl3 produced, we need to use the concept of stoichiometry and the balanced chemical equation for the reaction between Cr and Cl.
The balanced chemical equation for the reaction between Cr and Cl is:
2 Cr + 3 Cl2 -> 2 CrCl3
From the balanced equation, we can see that 2 moles of Cr react with 3 moles of Cl2 to produce 2 moles of CrCl3.
Step 1: Convert the given masses of Cr and Cl into moles.
To do this, we need to divide the given masses (in grams) by their respective molar masses.
Molar mass of Cr = 52 g/mol
Moles of Cr = mass of Cr / molar mass of Cr
Moles of Cr = 4.1 g / 52 g/mol
Molar mass of Cl = 35.5 g/mol
Moles of Cl = mass of Cl / molar mass of Cl
Moles of Cl = 9.3 g / 35.5 g/mol
Step 2: Determine the limiting reactant.
To determine the limiting reactant, we compare the moles of Cr to Cl using the stoichiometry of the balanced chemical equation.
From the balanced equation, we know that 2 moles of Cr react with 3 moles of Cl2. Therefore, the ratio of moles of Cr to moles of Cl2 is 2:3.
If the ratio of moles of Cr to moles of Cl2 is less than 2:3, then Cr is the limiting reactant. If the ratio is greater than 2:3, then Cl2 is the limiting reactant.
Step 3: Calculate the moles of CrCl3 produced.
Based on the stoichiometry of the balanced chemical equation, for every 2 moles of Cr, we obtain 2 moles of CrCl3.
Moles of CrCl3 = (moles of Cr / 2) x (moles of CrCl3 / moles of Cr)
Moles of CrCl3 = (moles of Cr / 2) x (2 moles of CrCl3 / 2 moles of Cr)
Step 4: Convert moles of CrCl3 to grams.
To do this, we multiply the moles of CrCl3 by the molar mass of CrCl3.
Molar mass of CrCl3 = molar mass of Cr + (3 x molar mass of Cl)
Molar mass of CrCl3 = 52 g/mol + (3 x 35.5 g/mol)
Mass of CrCl3 = Moles of CrCl3 x Molar mass of CrCl3
Finally, plug in the calculated values into the equation to get the mass of CrCl3 produced.