Chemistry help please?

A quantity of epsom salts, magnesium sulfate heptahydrate, MgSO₄•7H₂O, is heated until all the water is driven off. The sample loses 11.8 g in the process. What was the mass of the original sample?


Molar mass of MgSO₄ = (24.31 + 32.06 + 16.00×4) g/mol = 120.37 g/mol
Molar mass of MgSO₄•7H₂O = (120.37 + 1.008×14 + 16.00×7) g/mol = 246.48 g/mol

(Original mass of MgSO₄•7H₂O) : (Mass of H₂O) = 246.48 : (246.68 - 120.37)
(Original mass of MgSO₄•7H₂O) : (11.8 g) = 246.48 : 126.11
Original mass of MgSO₄•7H₂O = 11.8 g × 246.48 / 126.11 g = 23.1 g


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14. The process of manufacturing sulfuric acid begins with the burning of sulfur. What mass of sulfur would have to be burned in order to produce 1.00 kg of H₂SO₄? Assume that all of the sulfur ends up in the sulfuric acid.


Molar mass of S = 32.06 g/mol
Molar mass of H₂SO₄ = 1.008×2 + 32.06 + 16.00×4 = 98.08 g/mol

S →→ H₂SO₄
Mole ratio S : H₂SO₄ = 1 : 1

No. of moles of S = No. of moles of H₂SO₄ = (1000 g) / (98.08 g/mol)
Mass of S needed = [(1000 g) / (98.08 g/mol)] × (32.06 g/mol) = 327 g

The molar ratios are 1:7
mol(H2O) = 11.8 / 18 = 0.6555... moles ( Equivalent to '7'
mol(MgSO4) = 0.65555.... / 7 = 0.09365... moles (Equivalent to '1')

Mr (MgSO4)
Mg x 1 = 24 x 1 = 24
S x 1 = 32 x 1 = 32
O x 4 = 16 x 4 = 64
24 + 32 + 64 = 120
Hence
mass(MgSO4) = 0.09365 .. X 120 = 11,238 g (dehydrated MgSO4)
Original mass of 'heptahydrate' = 11.238 + 11.8 = 23.038g .

Sulphuric Acid
Reaction equations.
2S + 2O2 = 2SO2
2SO2 + O2 = 2SO3
2SO3 + 2H2O = 2H2SO4

NB 2 moles ratios of (H2SO4) is produced from 2 molar ratios of 'S'.

Mr(H2SO4)
H x 2 = 1 x 2 = 2
S x 1 = 32 x 1 = 32
O x 4 = 16 x 4 = 64
2 + 32 + 64 = 98
moles(H2SO4) = 1000 g / 98 = 10.204 moles ( Equivalent to '2' in the molar ratios).
moles(S) = 10.204 moles ( Also equivalent to '2' in the molar ratios).
mass(S) = 10.204 x 32 = 326.35 g
mass(S) = 0.32635 kg.

A.

1. For the first problem, you can consider this as a typical stoichiometric problem. Write the balanced equation for this reaction.

MgSO4-7H20 = MgSO4 + 7H2O

As you can see, the removal of water in a hydrate is a decomposition reaction. Keep in mind that the water in a hydrated salt is in stoichiometric ratio with its salt.

2. We are told in the problem that when MgSO4-7H2O was heated until all the water was driven off, the sample lost 11.8 g. Thus, this 11.8 grams is the mass of the water. We are asked the original mass of the salt. We can find the mass of the salt by relating it with the mass of water stoichiometrically.

3. Solving the problem using factor label method. I'm assuming you're familiar on how to calculate the molar mass of a compound and the mole concept.

11.8 g H2O x (1 mole H2O / 18 g H2O) x (1 mole MgSO4-7H2O / 7 moles H2O) x
(246.4746 g MgSO4-7H20 / 1 mole MgSO4-7H2O)
=23.08 g MgSO4-7H2O, original mass of the sample

B.

1. The second problem is another application of chemical stoichiometry. Let's analyze the mole ratio of the constituents of H2SO4.

1 mole H2SO4 = 2 moles H = 1 mole S = 4 moles O

It's pretty clear that we can arrive at desired mole ratio by inspecting the number of atoms in the constituent elements of the compound.

2. Using the factor label method

1.00 kg H2SO4 (1000 g H2SO4 / 1kg H2SO4) x (1 mole H2SO4 / 98.079 g H2SO4) x
(1 mole S / 1 mole H2SO4) x (32.065 g S / 1 mole S)
= 326.93 g S needed to be burned to produce 1.00 kg of H2SO4