A container can hold 65g of water. circle the conversion factor needed to find the mass of water that 5 identical containers can
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Answer:
Total pressure = 0,806 atm
Partial pressure of CH₄: 0,037 atm
Partial pressure of Cl₂: 0,396 atm
Partial pressure of CH₃Cl: 0,125 atm
Partial pressure of HCl: 0,125 atm
Partial pressure of Cl⁻: 0,125 atm
Explanation:
For the reaction:
2CH₄(g)+3Cl₂(g)⟶2CH₃Cl(g)+2HCl(g)+2Cl⁻(g)
295 mL≡ 0,295L of methane at STP are:
n = PV/RT
P = 1 atm; V = 0,295L; R = 0,082atmL/molK; T = 273K.
moles of methane: 0,0132 moles
For 725 mL of chlorine ≡ 0,725L
moles of chlorine at STP are: ≡ 0,0324 moles
For a complete reaction of 0,0132 moles of CH₄:
0,0132 mol CH₄× = <em>0,0198 moles</em>
The reaction reaches 77%, moles of Cl₂ that react are: 0,0198×77% = 0,0153 mol
As you have 0,0324 moles of Cl₂, moles that will not react are:
0,0324 - 0,0153 = <em>0,0171 mol Cl₂</em>
As the reaction reaches 77% completion, moles of CH₄ that react are:
0,0132×77% =<em> 0,0102 moles of CH₄ And the moles that don't react are </em><em>0,00300 mol</em>
Thus, moles of each compound are:
0,0102 moles of CH₄×= <em>0,0153 mol + 0,0171 mol = 0,0324 mol Cl₂</em>
0,0102 moles of CH₄×= <em>0,0102 mol CH₄</em>
0,0102 moles of CH₄×= <em>0,0102 mol HCl</em>
0,0102 moles of CH₄×= <em>0,0102 mol Cl⁻</em>
Total pressure using:
P = nRT/V
Where: n = 0,0102mol×3+0,0324mol + 0,0030mol = 0,0660mol; R = 0,082 atmL/molK; T = 298K; V = 2L
Total pressure = 0,806 atm
Partial pressure of CH₄: 0,037 atm
Partial pressure of Cl₂: 0,396 atm
Partial pressure of CH₃Cl: 0,125 atm
Partial pressure of HCl: 0,125 atm
Partial pressure of Cl⁻: 0,125 atm
<em>-To obtain partial pressure you change the moles for each compound-</em>
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I hope it helps!