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lesya [120]
3 years ago
8

How many grams of iron metal do you expect to be produced when 265 grams of an 84.5 percent by mass iron (II) nitrate solution r

eact with excess aluminum metal? Show all of the work needed to solve this problem.
2Al (s) + 3Fe(NO3)2 (aq) yields 3Fe (s) + 2Al(NO3)3 (aq)
Chemistry
1 answer:
Greeley [361]3 years ago
6 0
The balanced reaction is:

<span>2Al (s) + 3Fe(NO3)2 (aq) = 3Fe (s) + 2Al(NO3)3 (aq)
</span>
We are given the amount of <span>iron (II) nitrate solution with its purity. This will be the starting point of our calculation.

265 (.845) gram </span>iron (II) nitrate (1 mol Fe(NO3)2 / 179.85 g Fe(NO3)2) ( 3 mol Fe / 3 mol <span>Fe(NO3)2) (55.85 g Fe / 1 mol Fe) = 69.54 g Fe</span>
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Hello this is my 3 attempt, please someone answer this i'm begging you.
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Answer:

Explanation:

No worries, I got you :)

So for the first question, you need to use PV=nRT to find the n, or in other words, the number of moles. Then, you can find the molar mass since you know the grams and the moles

110 kPa / 101.3 = 1.085 atm (I converted it to atm so I can use the .08206 L atm/ k mol  for the rate)

550 ml / 1000 = .550 L (I converted mL to L in order to use the .08206 L atm/ k mol for the Rate)

28.5 c + 273 = 301.5 K (I converted C to K in order to use the .08206 L atm/ k mol for the Rate)

PV=nRT

(1.085) (.550 L) = n (0.08206) (301.5)

Divide the (0.08206) (301.5) to get n alone:

(1.085) (.550 L) / (0.08206) (301.5) = n

When I divided, I got n= .02412 moles, and since we have 1.88g , we divide the 1.88 by .02412 to get the molar mass (grams/mole)

77.94 g/mole is the molar mass

we know that there are 3 H's in the compound, so we do 3(1.008) and subtract 77.94 by what you get.

3 x 1.008 = 3.024 -----> 77.94-3.024 = 74.9

Now we look at the periodic table and try to find an element that has a molar mass of 74.9

Arsenic (As) has a molar mass of 74.922, which is close enough. Plus, Arsenic has a charge of 3, so it fits with the 3 hydrogens.

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A 0.1510 gram sample of a hydrocarbon produces 0.5008 gram CO2 and 0.1282 gram H2O in combustion analysis. Its
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In a combustion of a hydrocarbon compound, 2 reactions are happening per element:

C + O₂ → CO₂
2 H + 1/2 O₂ → H₂O

Thus, we can determine the amount of C and H from the masses of CO₂ and H₂O produced, respectively.

1.) Compute for the amount of C in the compound. The data you need to know are the following:
Molar mass of C = 12 g/mol
Molar mass of CO₂ = 44 g/mol
Solution:
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Compute for the amount of H in the compound. The data you need to know are the following:
Molar mass of H = 1 g/mol
Molar mass of H₂O = 18 g/mol
Solution:
0.1282 g H₂O*(1 mol H₂O/ 18 g)*(2 mol H/1 mol H₂O) = 0.014244 mol H
0.014244 mol H*(1 g/mol) = 0.014244 g H

The percent composition of pure hydrocarbon would be:
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Percent composition = (0.13658 g + 0.014244 g)/(<span>0.1510 g) * 100
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2. The empirical formula is determined by finding the ratio of the elements. From #1, the amounts of moles is:

Amount of C = 0.01138 mol
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Divide the least number between the two to each of their individual amounts:
C = 0.01138/0.01138 = 1
H = 0.014244/0.01138 = 1.25

The ratio should be a whole number. So, you multiple 4 to each of the ratios:
C = 1*4 = 4
H = 1.25*4 = 5

Thus, the empirical formula of the hydrocarbon is C₄H₅.

3. The molar mass of the empirical formula is

Molar mass = 4(12 g/mol) + 5(1 g/mol) = 53 g/mol
Divide this from the given molecular weight of 106 g/mol
106 g/mol / 53 g/mol = 2
Thus, you need to multiply 2 to the subscripts of the empirical formula.

Molecular Formula = C₈H₁₀

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