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3 years ago

1. What mass of H2O will be produced if 9.5 g of H2 reacts with 1.2 g of O2?

Chemistry

2 answers:

docker41 [41]3 years ago

8 0

1) The answer is: mass of water is 1.35 grams.

Balanced chemical reaction: 2H₂ + O₂ → 2H₂O.

m(O₂) = 1.2 g; mass of oxygen.

M(O₂) = 2 · 16 g/mol.

M = 32 g/mol, molar mass of oxygen.

n(O₂) = m(O₂) ÷ M(O₂).

n(O₂) = 1.2 g ÷ 32 g/mol.

n(O₂) = 0.0375 mol; amount of oxygen, limiting reactant.

m(H₂) = 9.5 g; mass of hydrogen.

M(H₂) = 2 g/mol, molar mass of hydrogen.

n(H₂) = m(H₂) ÷ M(H₂).

n(H₂) = 9.5 g ÷ 2 g/mol.

n(H₂) = 4.5 mol; amount of hydrogen.

From chemical reaction: n(O₂) : n(H₂O) = 1 : 2.

n(H₂O) = 0.0375 mol · 2.

n(H₂O) = 0.075 mol; amount of water.

m(H₂O) = n(H₂O) · M(H₂O).

m(H₂O) = 0.075 mol · 18 g/mol.

m(H₂O) = 1.35 g; mass of water.

2) The answer is: 3.05 grams of MgCl₂ is produced, the limiting reactant is Mg(OH)₂.

Balanced chemical reaction: 2HCl + Mg(OH)₂ → MgCl₂ + 2H₂O.

m(Mg(OH)₂) = 1.85 g; mass of magnesium hydroxide

n(Mg(OH)₂) = m(Mg(OH)₂) ÷ M(Mg(OH)₂).

n(Mg(OH)₂) = 1.85 g ÷ 58.32 g/mol.

n(Mg(OH)₂) = 0.032 mol; limiting reactant.

m(HCl) = 3.71 g; mass of hydrochloric acid.

n(HCl) = 3.71 g ÷ 36.46 g/mol.

n(HCl) = 0.102 mol; amount of hydrochloric acid.

From chemical reaction: n(Mg(OH)₂) : n(MgCl₂) = 1 : 1.

n(MgCl₂) =0.032 mol; amount of magnesium chloride.

m(MgCl₂) = n(MgCl₂) · M(MgCl₂).

m(MgCl₂) = 0.032 mol · 95.21 g/mol.

m(MgCl₂) = 3.05 g; mass of magnesium chloride.

3) The answer is: mass of potassium hydroxide is 8.1 grams.

Balanced chemical reaction: K₂O + H₂O → 2KOH.

m(K₂O) = 8.2 g; mass of potassium oxide.

n(K₂O) = m(K₂O) ÷ M(K₂O).

n(K₂O) = 8.2 g ÷ 94.2 g/mol.

n(K₂O) = 0.087 mol; amount of potassium oxide.

m(H₂O) = 1.3 g; mass of water.

n(H₂O) = 1.3 g ÷ 18 g/mol.

n(H₂O) = 0.072 mol; limiting reactant.

From chemical reaction: n(H₂O) : n(KOH) = 1 : 2.

n(KOH) = 2 · 0.072 mol.

n(KOH) = 0.144 mol.

m(KOH) = 0.144 mol · 56.1 g/mol.

m(KOH) = 8.1 g; mass of potassium hydroxide.

4) The answer is: mass of aluminum chloride is 16.66 grams.

Balanced chemical reaction: 2Al + 3Cl₂ → 2AlCl₃.

m(Al) = 8.1 g; mass of aluminium.

n(Al) = m(Al) ÷ M(Al).

n(Al) = 8.1 g ÷ 27 g/mol.

n(Al) = 0.3 mol.

V(Cl₂) = 4.2 L; volume of chlorine.

n(Cl₂) = 4.2 L ÷ 22.4 L/mol.

n(Cl₂) = 0.1875 mol; limiting reactant.

From chemical reaction: n(Cl₂) : n(AlCl₃) = 3 : 2.

n(AlCl₃) = 2 · 0.1875 mol ÷ 3.

n(AlCl₃) = 0.125 mol; amount of aluminium chloride.

m(AlCl₃) = 0.125 mol · 133.34 g/mol.

m(AlCl₃) = 16.66 g; mass of aluminium chloride.

erastovalidia [21]3 years ago

4 0

This is the number 3..

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$NH_{3}(g) + O_{2}(g) \: → NO_{2} + H_{2}O(g)$

__↑______↑

8.00 mol | 14.00 mol

________________

$NH_{3}(g) + O_{2}(g) \: → NO_{2} + H_{2}O(g)$

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To do this, create variables for the coefficients of each compound in the reaction respectively.

$a(NH_{3}(g)) + b(O_{2}(g)) → \\c(NO_{2}) + d(H_{2}O(g))$

Because to be balanced, the count of atoms in each element of the compound correspond to the coefficient of the variable in that compound so that the count of the left (reactant) side is set equal to the right (product) side.

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Reactant: 1a [N] Product: 1c.

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