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2 years ago

What is the number of mole equivalents of H2 produced by the reaction of 1 mole equivalent of Mg (m) with H2SO4 (aq)?

Chemistry

1 answer:

Inga [223]2 years ago

8 0

Answer:

We'll have 1.00 mol H2

Explanation:

Step 1: Data given

Number of moles of Mg = 1.00 moles

Step 2: The balanced equation

Mg(s) + H2SO4(aq) → MgSO4(aq) + H2(g)

Step 3: Calculate moles of H2

For 1 mol Mg we need 1 mol H2SO4 to produce 1 mol MgSO4 and 1 mol H2

We'll have 1.00 mol H2

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2 years ago

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3 years ago

If 5.57 g of Ag2O is sealed in a 75.0-mL tube filled with 760 torr of N2 gas at 28 ∘C, and the tube is heated to 310 ∘C, the Ag2

Snezhnost [94]

Answer: Total pressure = 7293.2 torr or 9.60 atm

Explanation:

Total pressure = partial pressure of nitrogen + partial pressure of oxygen

The partial pressure due to nitrogen is determined using the equation of Gay-Lussac's law: P₁/T₁=P₂/T₂

P₁ = 760 torr = 1atm, T₁ = 28∘C = (273+28)K = 301k, P₂ = ?, T₂ = 310∘C =(310+273)K = 583K

P₂ = P₁ T₂/ T₁

P₂ = 760 * 583 / 301 = 1472.03 torr

The pressure due to Oxygen gas produced is calculated thus:

Balanced equation of the decomposition of Ag₂O at s.t.p. is as follows;

2Ag₂O ----> 4Ag + O₂(g)

2 moles of Ag₂O produces 1 mole of O₂

molar mass of Ag₂O = (2*108 + 16)g = 232g/mol

molar volume of gas at s.t.p. = 22.4L

2*232g i.e. 464g of Ag₂O produces 22.4L of O₂

5.57g of Ag₂O will produce 5.57g*22.4L/464g = 0.269L or 269mL of O₂

Using the General gas equation P₁V₁/T₁=P₂V₂/T₂

P₁ = 1atm = 760 torr, V₁ = 269mL, T₁=273K, P₂ = ?, V₂= 75mL, T₂ = 583K

P₂ = P₁V₁T₂/V₂T₁

P₂ = 760*269*583 / 75*273

P₂ = 5821.17 torr

Total pressure = (1472.03 + 5821.17) torr

Total pressure = 7293.2 torr or 9.60 atm

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