Based on the following chemical equation, balance the equation an answer the following questions:
Fe2(S04)3 + NaCl ( FeCl3 + Na2S04
Calculate the amount in grams of NaCl needed to produce 75 g of FeCl3.
Calculate the amount of Fe2(S04)3 in grams produce 215 g Na2S04.
Calculate the number of formula units of FeCl3 produced with 95g of Fe2(S04)3.
Note that this reaction will not occur. I don't know if you made it up or a teacher gave it to you but it doesn't meet any of the criteria for a reaction to occur. However, for the puroses of answering your questions about stoichiometry we will assume it will occur. I will round all of my calculations and the molar masses so you will need to go through and recalculate each step. My final answer will be approximate only.
Fe2(S04)3 + 6NaCl ==> 2FeCl3 + 3Na2S04
a. moles in 75 g FeCl3 = grams/molar mass = 75/162 = 0.5
Convert to moles NaCl needed. Use the coefficients in the balanced equation.
0.5 moles FeCl3 x (6 mols NaCl/2 mols FeCl3) = 0.5 x 3 = 0.15
Grams NaCl = mols NaCl x molar mass NaCl = ?
b. c. b is worked the same as a. c follows the same guidelines.
Post your work if you get stuck.
Sure! Let's start by balancing the equation:
Fe2(S04)3 + 3NaCl --> 2FeCl3 + 3Na2S04
Now let's answer your questions:
1. Calculate the amount in grams of NaCl needed to produce 75 g of FeCl3.
To find this out, we need to use the balanced equation and stoichiometry. From the balanced equation, we see that 3 moles of NaCl react with 2 moles of FeCl3. We can calculate the moles of FeCl3:
75 g FeCl3 * (1 mole FeCl3 / atomic mass of FeCl3) = x moles of FeCl3
Now we can use the stoichiometry to find the moles of NaCl needed:
x moles FeCl3 * (3 moles NaCl / 2 moles FeCl3) = y moles NaCl
Finally, we can convert moles of NaCl to grams:
y moles NaCl * (atomic mass of NaCl / 1 mole NaCl) = amount in grams of NaCl needed
2. Calculate the amount of Fe2(S04)3 in grams produced from 215 g Na2S04.
Using the balanced equation, we can see that 3 moles of Na2S04 react with 1 mole of Fe2(S04)3. We can calculate the moles of Na2S04:
215 g Na2S04 * (1 mole Na2S04 / atomic mass of Na2S04) = x moles Na2S04
Now, using the stoichiometry:
x moles Na2S04 * (1 mole Fe2(S04)3 / 3 moles Na2S04) = y moles Fe2(S04)3
Finally, we can convert moles of Fe2(S04)3 to grams:
y moles Fe2(S04)3 * (atomic mass of Fe2(S04)3 / 1 mole Fe2(S04)3) = amount in grams of Fe2(S04)3
3. Calculate the number of formula units of FeCl3 produced using 95 g of Fe2(S04)3.
To calculate the number of formula units, we need to convert grams of Fe2(S04)3 to moles:
95 g Fe2(S04)3 * (1 mole Fe2(S04)3 / atomic mass of Fe2(S04)3) = x moles Fe2(S04)3
Now, using stoichiometry, we can find the number of formula units:
x moles Fe2(S04)3 * (2 moles FeCl3 / 1 mole Fe2(S04)3) * (Avogadro's number) = number of formula units of FeCl3 produced
I hope this helps! Let me know if you have any other questions, or if you'd like a joke to lighten the mood!
To balance the chemical equation Fe2(S04)3 + NaCl → FeCl3 + Na2S04, we need to ensure that the number of atoms on both sides of the equation is equal. Here is the balanced equation:
2 Fe2(SO4)3 + 6 NaCl → 3 FeCl3 + 3 Na2SO4
Now we can proceed to answer the questions.
1. Calculate the amount in grams of NaCl needed to produce 75 g of FeCl3:
- First, find the molar mass of FeCl3.
- FeCl3 consists of one iron atom (Fe) with a molar mass of 55.85 g/mol and three chlorine atoms (Cl) with a molar mass of 35.45 g/mol each.
- The total molar mass of FeCl3 is 55.85 g/mol + 3 * 35.45 g/mol = 162.2 g/mol.
- Since the balanced equation shows that 2 moles of FeCl3 are produced for every 6 moles of NaCl, we need to calculate the moles of FeCl3 needed first.
- Moles of FeCl3 = 75 g / 162.2 g/mol = 0.4621 mol.
- From the balanced equation, we know that 6 moles of NaCl react to produce 3 moles of FeCl3.
- So, moles of NaCl needed = 0.4621 mol * (6 mol NaCl / 3 mol FeCl3) = 0.9243 mol.
- Finally, calculate the grams of NaCl needed:
- Mass of NaCl = moles of NaCl needed * molar mass of NaCl.
- The molar mass of NaCl is 22.99 g/mol + 35.45 g/mol = 58.44 g/mol.
- Mass of NaCl = 0.9243 mol * 58.44 g/mol = 53.95 g.
- Therefore, the amount of NaCl needed to produce 75 g of FeCl3 is approximately 53.95 grams.
2. Calculate the amount of Fe2(S04)3 in grams to produce 215 g of Na2S04:
- First, find the molar mass of Na2SO4.
- Na2SO4 consists of two sodium atoms (Na) with a molar mass of 22.99 g/mol each, one sulfur atom (S) with a molar mass of 32.06 g/mol, and four oxygen atoms (O) with a molar mass of 16.00 g/mol each.
- The total molar mass of Na2SO4 is 2 * 22.99 g/mol + 32.06 g/mol + 4 * 16.00 g/mol = 142.04 g/mol.
- From the balanced equation, we know that 2 moles of Na2SO4 are produced for every 2 moles of Fe2(S04)3.
- So the moles of Na2SO4 produced equal the moles of Fe2(S04)3.
- Moles of Na2SO4 produced = 215 g / 142.04 g/mol = 1.5126 mol.
- Finally, calculate the grams of Fe2(S04)3:
- Mass of Fe2(S04)3 = moles of Fe2(S04)3 * molar mass of Fe2(S04)3.
- The molar mass of Fe2(S04)3 is (2 * 55.85 g/mol) + (3 * (32.06 g/mol + 4 * 16.00 g/mol)) = 399.88 g/mol.
- Mass of Fe2(S04)3 = 1.5126 mol * 399.88 g/mol = 606.24 g.
- Therefore, the amount of Fe2(S04)3 needed to produce 215 g of Na2SO4 is approximately 606.24 grams.
3. Calculate the number of formula units of FeCl3 produced with 95 g of Fe2(S04)3:
- First, calculate the moles of Fe2(S04)3 given its mass:
- Moles of Fe2(S04)3 = 95 g / 399.88 g/mol = 0.2377 mol.
- From the balanced equation, we know that 2 moles of Fe2(S04)3 react to produce 3 moles of FeCl3.
- So, moles of FeCl3 produced = 0.2377 mol * (3 mol FeCl3 / 2 mol Fe2(S04)3) = 0.3566 mol.
- Finally, calculate the number of formula units of FeCl3:
- Formula units of FeCl3 = moles of FeCl3 produced * Avogadro's number.
- Avogadro's number is approximately 6.022 x 10^23 formula units/mol.
- Formula units of FeCl3 = 0.3566 mol * 6.022 x 10^23 formu