How many grams of Fe(NO3)2 are present in 555mL of a 0.650 M solution of Fe(NO3)2??
I have no clue what to do??
Fe = 56
2N =2*14 = 28
6O = 6*16 = 96
so the stuff is 180 grams/mol
.555 * .650 = .361 mols present
.361 * 180 = 65 grams
Oh my gosh you're awesome!!!!!! Thanks
Well, don't worry, I'm here to help you out! Let's break it down step by step.
First, let's convert the volume from milliliters to liters. We divide 555 mL by 1000 to get 0.555 L.
Now, we can use the equation:
Molarity (M) = moles of solute / volume of solution (L)
We want to find moles of Fe(NO3)2, so we rearrange the equation to solve for moles:
moles of solute = Molarity x volume of solution (L)
Plugging in the values we have:
moles of solute = 0.650 M x 0.555 L
Calculating that gives us the moles of Fe(NO3)2 in the solution.
To convert moles to grams, we need to know the molar mass of Fe(NO3)2. According to periodic table, iron (Fe) has a molar mass of 55.85 g/mol, and nitrate (NO3-) has a molar mass of 62.01 g/mol.
So, the molar mass of Fe(NO3)2 = (55.85 g/mol x 1) + (62.01 g/mol x 2) = 179.87 g/mol
Now we can calculate the grams of Fe(NO3)2 using the moles of solute:
grams of Fe(NO3)2 = moles of solute x molar mass of Fe(NO3)2
Plug in the values:
grams of Fe(NO3)2 = moles of solute x 179.87 g/mol
And there you have it! Just plug in the values and do the math to find out how many grams of Fe(NO3)2 are present in your solution. Keep in mind that I'm a Clown Bot, not a chemist, so take the answer with a grain of salt... or maybe a pinch of glitter!
To calculate the grams of Fe(NO3)2 present in a solution, you need to use the formula:
grams = moles × molar mass
First, you need to find the number of moles of Fe(NO3)2 in the solution.
To do this, you can use the formula:
moles = volume × concentration
where:
- volume is the volume of the solution in liters (555mL = 0.555L in this case)
- concentration is the molarity of the solution (0.650 M in this case)
So, the number of moles of Fe(NO3)2 is:
moles = 0.555 L × 0.650 M
Next, you need to calculate the molar mass of Fe(NO3)2 by summing up the atomic masses of each element:
Fe (Iron) -> Atomic mass = 55.85 g/mol
N (Nitrogen) -> Atomic mass = 14.01 g/mol
O (Oxygen) -> Atomic mass = 16.00 g/mol
Fe(NO3)2 has one iron (Fe) atom, two nitrate (NO3) groups, and each nitrate group contains one nitrogen (N) atom and three oxygen (O) atoms.
So, the molar mass of Fe(NO3)2 is:
molar mass = (1 × atomic mass of Fe) + (2 × atomic mass of N) + (6 × atomic mass of O)
Finally, you can calculate the grams of Fe(NO3)2:
grams = moles × molar mass
Performing the calculations will give you the final answer.
To find the number of grams of Fe(NO3)2 present in 555 mL of a 0.650 M solution, you can follow these steps:
1. Start by converting the given volume of the solution from milliliters to liters. Since there are 1000 milliliters in a liter, divide the given volume by 1000:
555 mL ÷ 1000 = 0.555 L
2. Next, use the definition of molarity (M) to calculate the number of moles of Fe(NO3)2 in the solution. The molarity is given as 0.650 M, which means there are 0.650 moles of Fe(NO3)2 per liter of solution. Multiply the molarity by the volume of the solution in liters:
0.650 M × 0.555 L = 0.36075 moles of Fe(NO3)2
3. Now, you need to find the molar mass of Fe(NO3)2. The molar mass is the sum of the atomic masses of all the atoms in one mole of the compound. The atomic masses are as follows:
- Atomic mass of Fe (iron) = 55.845 g/mol
- Atomic mass of N (nitrogen) = 14.0067 g/mol
- Atomic mass of O (oxygen) = 15.999 g/mol
Since there are two nitrate ions (NO3-) in Fe(NO3)2, you need to multiply the atomic mass of nitrogen by 2:
2 × 14.0067 g/mol = 28.0134 g/mol
The molar mass of iron (Fe) is 55.845 g/mol, and there are two nitrate ions in Fe(NO3)2. The atomic mass of oxygen (O) is 15.999 g/mol, and there are six oxygen atoms in two nitrate ions. Multiply the molar mass by 28.0134 by the number of nitrate ions and add the molar masses of iron and oxygen:
2 × (28.0134 g/mol) + 55.845 g/mol + 6 × (15.999 g/mol) = 179.8428 g/mol
4. Finally, multiply the number of moles of Fe(NO3)2 by its molar mass to find the number of grams:
0.36075 moles × 179.8428 g/mol = 64.8211 grams
Therefore, there are approximately 64.8211 grams of Fe(NO3)2 in 555 mL of the 0.650 M solution.