Answer:
116.3 kJ
Step-by-step explanation:
Three heat transfers are involved
q = Heat to warm ice + heat to melt ice + heat to warm water + heat to evaporate water + heat to warm steam
q = q₁ + q₂ + q₃ + q₄ + q₅
q = mC₁ΔT₁ + mΔH_fus + mC₃ΔT₃ + mΔH_vap + mC₅ΔT₅
<em>Step 1</em>: Calculate q₁
m = 37.0 g
C₁ = 2.010 J·°C⁻¹g⁻¹
ΔT₁ = T_f – T_i
ΔT₁ = 0.0 – (-10.0)
ΔT₁ = 10.0 °C
q₁ = 37.0 × 2.010 × 10.0
q₁= 743.7 J
q₁= 0.7437 kJ
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<em>Step 2</em>. Calculate q₂
ΔH_fus = 334 J/g
q₂ = 37.0 × 334
q₂ = 12 360 J
q₂ = 12.36 kJ
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Step 3: Calculate q₃
C₃ = 4.179 J·°C⁻¹g⁻¹
ΔT₃ = T_f – T_i
ΔT₃ = 100 – 0
ΔT₃ = 100 °C
q₃ = 37.0 × 4.179 × 100
q₃ = 15 460 J
q₃ = 15.46 kJ
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<em>Step 4</em>. Calculate q₄
ΔH_vap = 2260 J/g
q₄ = 37.0 × 2260
q₄ = 83 620 J
q₄ = 83.62 kJ
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<em>Step 5</em>. Calculate q₅
C¬₅ = 2.010 J·°C⁻¹g⁻¹
ΔT₅ = T_f – T_i
ΔT₅ = 155.0 – 1000
ΔT₅ = 55.0 °C
q₅ = 37.0 × 2.010 × 55
q₅ = 4090 J
q₅ = 4.090 kJ
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Step 6. Calculate q
q = 0.7437 + 12.36 + 15.46 + 83.62 + 4.090
q = 116.3 kJ
The heat required is 116.3 kJ.
Answer:
68.1% is percent yield of the reaction
Explanation:
The reaction of methane with oxygen is:
CH₄ + 2O₂ → CO₂ + 2H₂O
<em>Where 2 moles of oxygen react per mole of CH₄</em>
<em />
Percent yield is:
Actual yield (28.2g CO₂) / Theoretical yield * 100
To solve this question we need to find theoretical yield finding limiting reactant :
<em>Moles CH₄:</em>
15.1g CH₄ * (1mol / 16.04g) = 0.9414 moles
<em>Moles O₂:</em>
81.2g * (1mol / 32g) = 2.54 moles
For a complete reaction of 0.9414 moles of CH₄ are needed:
0.9414 moles CH₄ * (2 mol O₂ / 1mol CH₄) = 1.88 moles of O₂. As there are 2.54 moles, O₂ is in excess and <em>CH₄ is limiting reactant</em>
In theoretical yield, the moles of methane added = Moles of CO₂ produced. That is 0.9414 moles CO₂. In grams = Theoretical yield:
0.9414 moles CO₂ * (44.01g / mol) = 41.43g CO₂
Percent yield: 28.2g CO₂ / 41.43g CO₂ * 100=
<h3>68.1% is percent yield of the reaction</h3>
H2S04 is the answer to this question
