if someone could help me with my chem with work that would be great?

9.
a.
2H₂(g) + O₂(g) → 2H₂O(l)
According to the equation, mole ratio H₂ : O₂ = 2 : 1
If 2.3 mol O₂ completely reacts, H₂ needed = (2.3 mol) × (2/1) = 4.6 mol < 5.0 mol
H₂ is in excess, and thus O₂ is the limiting reagent.

b.
2Al(s) + 3Cl₂(g) → 2AlCl₃(s)
According to the equation, mole ratio Al : Cl₂ = 2 : 3
If 8.0 mol Cl₂ completely reacts, Al needed = (8.0 mol) × (2/3) = 5.3 mol < 5.4 mol
Al is in excess, and thus O₂ is the limiting reagent.

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10.
For the first equation: C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(l)
Mole ratio C₂H₄ : O₂ : H₂O = 1 : 3 : 2

a.
If 6.3 mol O₂ completely reacts, C₂H₄ needed = (6.3 mol) × (1/3) = 2.1 mol < 2.7 mol
C₂H₄ is in excess, and thus O₂ in the limiting reagent.

b.
Moles of H₂O produced = (6.3 mol) × (2/3) = 4.2 mol

c.
Moles of C₂H₄ in excess = (2.7 mol) - (2.1 mol) = 0.6 mol

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For the first equation: C₂H₄(g) + 2O₂(g) → 2CO(g) + 2H₂O(l)
Mole ratio C₂H₄ : O₂ : H₂O = 1 : 2 : 2

a.
If 2.70 mol C₂H₄ completely reacts, O₂ needed = (2.7 mol) × (2/1) = 5.4 mol < 6.3 mol
O₂ is in excess, and thus C₂H₄ in the limiting reagent.

b.
Moles of H₂O produced = (2.7 mol) × (2/1) = 5.4 mol

c.
Moles of O₂ in excess = (6.3 mol) - (5.4 mol) = 0.9 mol

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11.
a.
Molar mass of N₂H₄ = (14.0×2 + 1.0×4) g/mol = 32.0 g/mol
Molar mass of O₂ = 16.0×2 g/mol = 32.0 g/mol

Initial moles of N₂H₄ reacted = (1000 g) / (32.0 g/mol) = 31.25 mol
Initial moles of O₂ reacted = (1000 g) / (32.0 g/mol) = 31.25 mol

According to the equation: N₂H₄(l) + O₂(g) → N₂(g) + 2H₂O(g)
Mole ratio N₂H₄ : O₂ : N₂ = 1 : 1 : 1
If 31.25 mol N₂H₄ completely reacts, moles of O₂ needed = 31.25 mol = Initial moles of O₂
Hence, both N₂H₄ and O₂ completely react.
Moles of N₂ formed = 31.25 mol

Molar volume of N₂ gas at STP = 22.4 L/mol
Volume of N₂ formed = (31.25 mol) × (22.4 L/mol) = 700 L

b.
As both N₂H₄ and O₂ completely react, no N₂O₄ or O₂ remain after the reaction.

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12.
Molar mass of CuO = (63.5 + 16.0) g/mol = 79.5 g/mol
Molar mass of CH₄ = (12.0 + 1.0×4) g/mol = 16.0 g/mol

Initial moles of CuO = (30.0 g) / (79.5 g/mol) = 0.3774 mol
Initial moles of CH₄ = (20.0 g) / (16.0 g/mol) = 1.25 mol

According to the balanced equation: 2CuO(s) + CH₄(g) → 2H₂O(l) + 2Cu(s) + CO₂(g)
Mole ratio CuO : CH₄ : Cu = 2 : 1 : 2
If 0.3774 mol CuO completely reacts, moles CH₄ needed = (0.3774 mol) × (1/2) = 0.1887 mol < 1.25 mol
Hence, CH₄ is in excess, and CuO is the limiting reagent.
Moles of Cu produced = (0.3774 mol) × (2/2) = 0.3774 mol

Molar mass of Cu = 63.5 g/mol
Maximum mass of Cu produced = (0.3774 mol) × (63.5 g/mol) = 24.0 g (to 3 sig. fig.)