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gulaghasi [49]
3 years ago
7

How many grams of Sulfuric Acid are needed to produce 57.18 g of Lead (IV) Sulfate when being neutralized by a sufficient amount

of Lead (IV) Hydroxide? *
Chemistry
1 answer:
KiRa [710]3 years ago
7 0

Answer:

40.72g of sulfuric acid are needed

Explanation:

When sulfuric acid, H₂SO₄, is neutralized by lead (IV) hydroxide, Pb(OH)₄, Lead (IV) sulfate, Pb(SO₄)₂ and water as follows:

2 H₂SO₄ + Pb(OH)₄ → Pb(SO₄)₂ + 4H₂O

To solve this question we must find the moles of 57.18g of Pb(SO₄)₂. As 2 moles of H₂SO₄ produce 1mol Pb(SO₄)₂ we can find the moles of H₂SO₄ and its mass as follows:

<em>Moles Pb(SO₄)₂ -Molar mass: 275.23 g/mol-</em>

57.18g * (1mol / 275.23g) = 0.2078 moles Pb(SO₄)₂

<em>Moles H₂SO₄:</em>

0.2078 moles Pb(SO₄)₂ * (2mol H₂SO₄ / 1mol Pb(SO₄)₂) = 0.4155 moles H₂SO₄

<em>Mass H₂SO₄ -Molar mass: 98g/mol-</em>

0.4155 moles H₂SO₄ * (98g / mol) =

<h3>40.72g of sulfuric acid are needed</h3>
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Find the amount of heat energy needed to convert 400 grams of ice at -38°C to steam at 160°C.
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The amount of heat energy needed to convert 400 g of ice at -38 °C to steam at 160 °C is 1.28×10⁶ J (Option D)

<h3>How to determine the heat required change the temperature from –38 °C to 0 °C </h3>
  • Mass (M) = 400 g = 400 / 1000 = 0.4 Kg
  • Initial temperature (T₁) = –25 °C
  • Final temperature (T₂) = 0 °
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  • Specific heat capacity (C) = 2050 J/(kg·°C)
  • Heat (Q₁) =?

Q = MCΔT

Q₁ = 0.4 × 2050 × 38

Q₁ = 31160 J

<h3>How to determine the heat required to melt the ice at 0 °C</h3>
  • Mass (m) = 0.4 Kg
  • Latent heat of fusion (L) = 334 KJ/Kg = 334 × 1000 = 334000 J/Kg
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Q = mL

Q₂ = 0.4 × 334000

Q₂ = 133600 J

<h3>How to determine the heat required to change the temperature from 0 °C to 100 °C </h3>
  • Mass (M) = 0.4 Kg
  • Initial temperature (T₁) = 0 °C
  • Final temperature (T₂) = 100 °C
  • Change in temperature (ΔT) = 100 – 0 = 100 °C
  • Specific heat capacity (C) = 4180 J/(kg·°C)
  • Heat (Q₃) =?

Q = MCΔT

Q₃ = 0.4 × 4180 × 100

Q₃ = 167200 J

<h3>How to determine the heat required to vaporize the water at 100 °C</h3>
  • Mass (m) = 0.4 Kg
  • Latent heat of vaporisation (Hv) = 2260 KJ/Kg = 2260 × 1000 = 2260000 J/Kg
  • Heat (Q₄) =?

Q = mHv

Q₄ = 0.4 × 2260000

Q₄ = 904000 J

<h3>How to determine the heat required to change the temperature from 100 °C to 160 °C </h3>
  • Mass (M) = 0.4 Kg
  • Initial temperature (T₁) = 100 °C
  • Final temperature (T₂) = 160 °C
  • Change in temperature (ΔT) = 160 – 100 = 60 °C
  • Specific heat capacity (C) = 1996 J/(kg·°C)
  • Heat (Q₅) =?

Q = MCΔT

Q₅ = 0.4 × 1996 × 60

Q₅ = 47904 J

<h3>How to determine the heat required to change the temperature from –38 °C to 160 °C</h3>
  • Heat for –38 °C to 0°C (Q₁) = 31160 J
  • Heat for melting (Q₂) = 133600 J
  • Heat for 0 °C to 100 °C (Q₃) = 167200 J
  • Heat for vaporization (Q₄) = 904000 J
  • Heat for 100 °C to 160 °C (Q₅) = 47904 J
  • Heat for –38 °C to 160 °C (Qₜ) =?

Qₜ = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

Qₜ = 31160 + 133600 + 167200 + 904000 + 47904

Qₜ = 1.28×10⁶ J

Learn more about heat transfer:

brainly.com/question/10286596

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