Sovi.AI - AI Math Tutor

Scan to solve math questions
ENFRDEESJAZH-CNZH-TW

QUESTION IMAGE

identify and balance - 17 1) h₃po₄ → h₄p₂o₇ + h₂o 2) ch₂ + o₂ → h₂o + c…

Question

identify and balance - 17

  1. h₃po₄ → h₄p₂o₇ + h₂o
  2. ch₂ + o₂ → h₂o + co₂
  3. agno₃ + zn → ag + zn(no₃)₂
  4. fecl₃ + nh₄oh → fe(oh)₃ + nh₄cl
  5. n₂ + h₂ → nh₃
  6. bacl₂ + al₂(so₄)₃ → baso₄ + alcl₃
  7. n₂ + o₂ → n₂o
  8. h₃aso₄ → as₂o₅ + h₂o
  9. cl₂ + nabr → nacl + br₂
  10. c₇h₁₆ + o₂ → co₂ + h₂o
  11. h₃po₄ + koh → k₃po₄ + h₂o
  12. zn + hcl → zncl₂ + h₂
  13. so₃ → s₈ + o₂
  14. k + br₂ → kbr
  15. c₂h₅oh + o₂ → co + h₂o

Explanation:

Step1: Balance equation 1: \( \ce{H3PO4 -> H4P2O7 + H2O} \)

Count atoms: P: 1 left, 2 right. Multiply \( \ce{H3PO4} \) by 2: \( 2\ce{H3PO4} \). Now H: 6 left, \( 4 + 2 = 6 \) right. O: \( 2\times4 = 8 \) left, \( 7 + 1 = 8 \) right. So balanced: \( 2\ce{H3PO4} = \ce{H4P2O7} + \ce{H2O} \)

Step2: Balance equation 2: \( \ce{CH2 + O2 -> H2O + CO2} \)

Carbon: 1 left, 1 right. Hydrogen: 2 left, 2 right. Oxygen: 2 left, \( 1 + 2 = 3 \) right. Multiply \( \ce{O2} \) by \( \frac{3}{2} \), but use integers. Multiply all by 2: \( 2\ce{CH2} + 3\ce{O2} = 2\ce{H2O} + 2\ce{CO2} \)

Step3: Balance equation 3: \( \ce{AgNO3 + Zn -> Ag + Zn(NO3)2} \)

\( \ce{NO3^-} \): 1 left, 2 right. Multiply \( \ce{AgNO3} \) by 2: \( 2\ce{AgNO3} + \ce{Zn} = 2\ce{Ag} + \ce{Zn(NO3)2} \)

Step4: Balance equation 4: \( \ce{FeCl3 + NH4OH -> Fe(OH)3 + NH4Cl} \)

\( \ce{Cl^-} \): 3 left, 1 right. Multiply \( \ce{NH4Cl} \) by 3. \( \ce{OH^-} \): 1 left, 3 right. Multiply \( \ce{NH4OH} \) by 3: \( \ce{FeCl3} + 3\ce{NH4OH} = \ce{Fe(OH)3} + 3\ce{NH4Cl} \)

Step5: Balance equation 5: \( \ce{N2 + H2 -> NH3} \)

N: 2 left, 1 right. Multiply \( \ce{NH3} \) by 2. H: 2 left, 6 right. Multiply \( \ce{H2} \) by 3: \( \ce{N2} + 3\ce{H2} = 2\ce{NH3} \)

Step6: Balance equation 6: \( \ce{BaCl2 + Al2(SO4)3 -> BaSO4 + AlCl3} \)

\( \ce{SO4^{2-}} \): 3 left, 1 right. Multiply \( \ce{BaSO4} \) by 3. \( \ce{Ba^{2+}} \): 1 left, 3 right. Multiply \( \ce{BaCl2} \) by 3. \( \ce{Cl^-} \): \( 3\times2 = 6 \) left, 3 right. Multiply \( \ce{AlCl3} \) by 2. \( \ce{Al^{3+}} \): 2 left, 2 right. So: \( 3\ce{BaCl2} + \ce{Al2(SO4)3} = 3\ce{BaSO4} + 2\ce{AlCl3} \)

Step7: Balance equation 7: \( \ce{N2 + O2 -> N2O} \) (assuming typo, product \( \ce{N2O} \))

O: 2 left, 1 right. Multiply \( \ce{N2O} \) by 2. N: 2 left, 4 right. Multiply \( \ce{N2} \) by 2: \( 2\ce{N2} + \ce{O2} = 2\ce{N2O} \) (if product is \( \ce{NO} \), different, but based on likely typo)

Step8: Balance equation 8: \( \ce{H3AsO4 -> As2O5 + H2O} \)

As: 1 left, 2 right. Multiply \( \ce{H3AsO4} \) by 2. H: 6 left, 2 right. Multiply \( \ce{H2O} \) by 3. O: \( 2\times4 = 8 \) left, \( 5 + 3 = 8 \) right: \( 2\ce{H3AsO4} = \ce{As2O5} + 3\ce{H2O} \)

Step9: Balance equation 9: \( \ce{Cl2 + NaBr -> NaCl + Br2} \)

Br: 1 left, 2 right. Multiply \( \ce{NaBr} \) by 2. Cl: 2 left, 2 right (after multiplying \( \ce{NaCl} \) by 2): \( \ce{Cl2} + 2\ce{NaBr} = 2\ce{NaCl} + \ce{Br2} \)

Step10: Balance equation 10: \( \ce{C7H16 + O2 -> CO2 + H2O} \)

C: 7 left, 1 right. Multiply \( \ce{CO2} \) by 7. H: 16 left, 2 right. Multiply \( \ce{H2O} \) by 8. O: 2 left, \( 7\times2 + 8 = 22 \) right. Multiply \( \ce{O2} \) by 11: \( \ce{C7H16} + 11\ce{O2} = 7\ce{CO2} + 8\ce{H2O} \)

Step11: Balance equation 11: \( \ce{H3PO4 + KOH -> K3PO4 + H2O} \)

\( \ce{K^+} \): 1 left, 3 right. Multiply \( \ce{KOH} \) by 3. H: \( 3 + 3 = 6 \) left, 2 right. Multiply \( \ce{H2O} \) by 3. O: \( 4 + 3 = 7 \) left, \( 4 + 3 = 7 \) right: \( \ce{H3PO4} + 3\ce{KOH} = \ce{K3PO4} + 3\ce{H2O} \)

Step12: Balance equation 12: \( \ce{Zn + HCl -> ZnCl2 + H2} \)

Cl: 1 left, 2 right. Multiply \( \ce{HCl} \) by 2. H: 2 left, 2 right. Zn: 1 left, 1 right: \( \ce{Zn} + 2\ce{HCl} = \ce{ZnCl2} + \ce{H2} \)

Step13: Balance equation 13: \( \ce{SO3 -> S8 + O2} \)

S: 1 left, 8 right. Multiply \( \ce{SO3} \) by 8. O: \( 8\times3 = 24 \) left, 2 right. Multiply \( \ce{O2} \) by 12: \( 8\ce{SO3} = \ce{S8} + 12\ce{O2} \)

Step14: Balance equation 14: \( \ce{K + Br2 -> KBr} \)

Br: 2 left, 1 right. Multiply \( \ce{KBr} \) by 2. K: 1 left, 2 right. Multiply \( \ce{K} \) by 2: \( 2\c…

Answer:

  1. \( 2\ce{H3PO4} = \ce{H4P2O7} + \ce{H2O} \)
  2. \( 2\ce{CH2} + 3\ce{O2} = 2\ce{H2O} + 2\ce{CO2} \)
  3. \( 2\ce{AgNO3} + \ce{Zn} = 2\ce{Ag} + \ce{Zn(NO3)2} \)
  4. \( \ce{FeCl3} + 3\ce{NH4OH} = \ce{Fe(OH)3} + 3\ce{NH4Cl} \)
  5. \( \ce{N2} + 3\ce{H2} = 2\ce{NH3} \)
  6. \( 3\ce{BaCl2} + \ce{Al2(SO4)3} = 3\ce{BaSO4} + 2\ce{AlCl3} \)
  7. \( 2\ce{N2} + \ce{O2} = 2\ce{N2O} \) (or adjust for correct product)
  8. \( 2\ce{H3AsO4} = \ce{As2O5} + 3\ce{H2O} \)
  9. \( \ce{Cl2} + 2\ce{NaBr} = 2\ce{NaCl} + \ce{Br2} \)
  10. \( \ce{C7H16} + 11\ce{O2} = 7\ce{CO2} + 8\ce{H2O} \)
  11. \( \ce{H3PO4} + 3\ce{KOH} = \ce{K3PO4} + 3\ce{H2O} \)
  12. \( \ce{Zn} + 2\ce{HCl} = \ce{ZnCl2} + \ce{H2} \)
  13. \( 8\ce{SO3} = \ce{S8} + 12\ce{O2} \)
  14. \( 2\ce{K} + \ce{Br2} = 2\ce{KBr} \)
  15. \( \ce{C2H5OH} + 2\ce{O2} = 2\ce{CO} + 3\ce{H2O} \) (or for complete combustion: \( \ce{C2H5OH} + 3\ce{O2} = 2\ce{CO2} + 3\ce{H2O} \))