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swat32
3 years ago
12

Write the complete ionic equation and the Net ionic equation for: K2C2O4(aq)+Pb(OH)2(aq) -> 2KOH(aq)+PbC2O4(s)

Chemistry
1 answer:
Kipish [7]3 years ago
7 0
Molecular chemical equation: 
K₂C₂O₄(aq)+Pb(OH)₂(aq) → 2KOH(aq) + PbC₂O₄<span>(s).
Ionic equation:
2K</span>⁺(aq) + C₂O₄²⁻(aq) + Pb²⁺(aq) + 2OH⁻(aq) → 2K⁺(aq) +2OH⁻(aq)+PbC₂O₄(s)
Net ionic eqation:
C₂O₄²⁻(aq) + Pb²⁺(aq) → PbC₂O₄(s).
s is solid, do not dissolve in water, potassiom hydroxide is trong base and dissolves in water.
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A 1.0857 gram pure sample of a compound containing only carbon, hydrogen, and oxygen was burned in excess oxygen gas. 2.190 g of
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Answer:

  • C₂ H₄ O

Explanation:

<u>1) Mass of carbon (C) in 2.190 g of carbon dioxide (CO₂)</u>

  • atomic mass of C: 12.0107 g/mol
  • molar mass of CO₂: 44.01 g/mol
  • Set a proportion: 12.0107 g of C / 44.01 g of CO₂ = x / 2.190 g of CO₂
  • Solve for x:

         x = (12.0107 g of C / 44.01 g of CO₂ ) × 2.190 g of CO₂ = 0.59767 g of C

<u />

<u>2) Mass of hydrogen (H) in 0.930 g of water (H₂O)</u>

  • atomic mass of H: 1.00784 g/mol
  • molar mass of H₂O: 18.01528 g/mol
  • proportion: 2 × 1.00784 g of H / 18.01528 g of H₂O = x / 0.930 g of H₂O
  • Solve for x:

        x = ( 2 × 1.00784 g of H / 18.01528 g of H₂O) × 0.930 g of H₂O = 0.10406 g of H

<u>3) Mass of oxygen (O) in 1.0857 g of pure sample</u>

  • Mass of O = mass of pure sample - mass of C - mass of H
  • Mass of O = 1.0857 g - 0.59767 g - 0.10406 = 0.38397 g O

Round to four decimals: Mass of O = 0.3840 g

<u>4) Mole calculations</u>

Divide the mass in grams of each element by its atomic mass:

  • C: 0.59767 g / 12.0107 g/mol = 0.04976 mol
  • H: 0.10406 g / 1.00784 g/mol = 0.10325 mol
  • O: 0.3840 g / 15.999 g/mol = 0.02400 mol

<u>5) Divide every amount by the smallest value (to find the mole ratios)</u>

  • C: 0.04976 mol / 0.02400 mol = 2.07 ≈ 2
  • H: 0.10325 mol / 0.02400 mol = 4.3 ≈ 4
  • O: 0.02400 mol / 0.02400 mol = 1

Thus the mole ratio is 2 : 4 : 1, and the empirical formula is:

  • <u>C₂ H₄ O </u>← answer
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