P L E A S E help! explain and give answers on how to do moles for chemistry

1. Silver acetate → 166.87 g/m, 2.50 moles, 417.2 g, 1.50×10²⁴ particles

2. Glucose → 180 g/mol, 1.8 moles, 324 g, 1.08×10²⁴ particles

3. Lead sulfide → 239.26 g/m, 0.522 moles, 125 g, 3.14×10²³ particles

4. Iron (III) Chloride → 162.2 g/m, 0.390 moles, 63.3 g,  2.35×10²³ particles

5. Aluminum sulfate → 342.14 g/m, 1.56 mol, 533.7 g, 9.39×10²³ particles

6. Caffeine → 194 g/m, 7.17 moles, 1392 g, 4.32×10²⁴ particles

13.1, 83.9 L of N₂

13.2, 7.59 L of C₂H₆

13.3 232.8 L of SO₃

Explanation:

Molar mass Ag + 2 molar mass C + 3 molar mass H + 2 molar mass O ⇒ 107.87 g/m + 2 . 12 g/m + 3 . 1 g/m + 2 . 16 g/m =  166.87 g/m (molar mass)

If we have 2.50 moles , we have → 2.5 mol . 166.87 g/m = 417.2 g

1 mol of salt has 6.02×10²³ representative particles

2.5 moles of salt must have (2.5 . 6.02×10²³) / 1 = 1.50×10²⁴ particles

Molar mass → 180 g/mol

Let's convert the mass to moles → 324 g . 1 mol / 180 g = 1.8 moles

To determine number of particles → 1.8 mol . 6.02×10²³ particles / 1 mol =

1.08×10²⁴ particles

207.2 g/m + 32.06 g/m = 239.26 g/m

Mass to moles → 125 g . 1 mol / 239.26 g = 0.522 moles

0.522 moles . 6.02×10²³ particles / 1 mol = 3.14×10²³ representative particles.

Molar mass = Molar mass Fe + 3 Molar mass Cl

55.85 g/m + 3 . 35.45 g/m = 162.2 g/m

With the representative particles, we determine the moles.

2.35×10²³ particles . 1 mol / 6.02×10²³ particles = 0.390 moles

0.390 mol . 162.2 g / 1 mol = 63.3 g

Molar mass → 2 molar mass Al + 3 molar mass S + 12 molar mass O

2. 26.98 g/m + 3 . 32.06 g/m + 12 . 16 g/m = 342.14 g/m

We determine mass → 342.14 g /m . 1.56 mol = 533.7 g

1.56 moles . 6.02×10²³ particles / 1 mol = 9.39×10²³ particles

Molar mass → 8 . 12 g/m + 10 . 1 g/m + 4 . 14 g/m + 2. 16 g/m = 194 g/m

We determine the moles, by the representative particles:

4.32×10²⁴ particles . 1mol / 6.02×10²³ = 7.17 moles

We convert the moles to mass, 7.17 mol . 194g / 1 mol = 1392 g

Excercise 13:

1. P . V = n .  R . T

V = n . R . T / P → 3.75 mol . 0.082L.atm /mol.K . 273K / 1 atm = 83.9L

2. P .V = n . R . T

V =  n . R . T / P → 0.339 mol . 0.082L.atm /mol.K . 273K / 1 atm = 7.59L

3. We determine the moles of SO₃ → mass / molar mass

835 g / 80.06 g/m = 10.4 moles

V =  n . R . T / P → 10.4 mol . 0.082L.atm /mol.K . 273K / 1 atm = 232.8L

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

its going to be mutualism cuz they both benefit from the relationship.

~batmans wife dun dun