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Balance the following chemical equations: Na(s) + H2O(l) → NaOH(aq) + H2(g) NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + H2O(l) CuCO3(aq) + HNO3(aq) → Cu(NO3)2(aq) + CO2(g) + H2O(l) H2S(g) + 3O2(g) → SO2(g) + H2O(l) C2H6(g) + O2(g) → CO2(g) + H2O(l)

Chemistry

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Haya njirudionisia69, tuendelee na maswali ya kusawazisha milinganyo.

12. Sawazisha milinganyo ifuatayo:

Part b· 3 steps

Na(s)+H2O(l)→NaOH(aq)+H2(g)\text{Na(s)} + \text{H}_2\text{O(l)} \rightarrow \text{NaOH(aq)} + \text{H}_2\text{(g)}

  1. Angalia idadi ya atomi za kila elementi pande zote mbili.

    Upande wa kushoto: Na=1\text{Na}=1, H=2\text{H}=2, O=1\text{O}=1.
    Upande wa kulia: Na=1\text{Na}=1, H=1+2=3\text{H}=1+2=3, O=1\text{O}=1.

  2. Sawazisha atomi za hidrojeni na oksijeni.

    Weka koefishienti 2 mbele ya H2O\text{H}_2\text{O} na 2 mbele ya NaOH\text{NaOH} na 2 mbele ya Na\text{Na}.

    2Na(s)+2H2O(l)→2NaOH(aq)+H2(g)2\text{Na(s)} + 2\text{H}_2\text{O(l)} \rightarrow 2\text{NaOH(aq)} + \text{H}_2\text{(g)}
  3. Angalia tena idadi ya atomi za kila elementi.

    Upande wa kushoto: Na=2\text{Na}=2, H=2×2=4\text{H}=2 \times 2 = 4, O=2\text{O}=2.
    Upande wa kulia: Na=2\text{Na}=2, H=2+2=4\text{H}=2+2=4, O=2\text{O}=2.
    Mlinganyo sasa umesawazishwa.

    Jibu: 2Na(s)+2H2O(l)→2NaOH(aq)+H2(g)\boxed{2\text{Na(s)} + 2\text{H}_2\text{O(l)} \rightarrow 2\text{NaOH(aq)} + \text{H}_2\text{(g)}}

Part c· 4 steps

NaOH(aq)+H2SO4(aq)→Na2SO4(aq)+H2O(l)\text{NaOH(aq)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{Na}_2\text{SO}_4\text{(aq)} + \text{H}_2\text{O(l)}

  1. Angalia idadi ya atomi za kila elementi pande zote mbili.

    Upande wa kushoto: Na=1\text{Na}=1, O=1+4=5\text{O}=1+4=5, H=1+2=3\text{H}=1+2=3, S=1\text{S}=1.
    Upande wa kulia: Na=2\text{Na}=2, O=4+1=5\text{O}=4+1=5, H=2\text{H}=2, S=1\text{S}=1.

  2. Sawazisha atomi za sodiamu (Na\text{Na}).

    Kuna atomi 2 za Na\text{Na} upande wa kulia na 1 upande wa kushoto. Weka koefishienti 2 mbele ya NaOH\text{NaOH}.

    2NaOH(aq)+H2SO4(aq)→Na2SO4(aq)+H2O(l)2\text{NaOH(aq)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{Na}_2\text{SO}_4\text{(aq)} + \text{H}_2\text{O(l)}
  3. Sawazisha atomi za hidrojeni (H\text{H}) na oksijeni (O\text{O}).

    Sasa kuna 2×1+2=42 \times 1 + 2 = 4 atomi za H\text{H} upande wa kushoto na 2 upande wa kulia. Weka koefishienti 2 mbele ya H2O\text{H}_2\text{O}.

    2NaOH(aq)+H2SO4(aq)→Na2SO4(aq)+2H2O(l)2\text{NaOH(aq)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{Na}_2\text{SO}_4\text{(aq)} + 2\text{H}_2\text{O(l)}
  4. Angalia tena idadi ya atomi za kila elementi.

    Upande wa kushoto: Na=2\text{Na}=2, O=2+4=6\text{O}=2+4=6, H=2+2=4\text{H}=2+2=4, S=1\text{S}=1.
    Upande wa kulia: Na=2\text{Na}=2, O=4+2=6\text{O}=4+2=6, H=2×2=4\text{H}=2 \times 2 = 4, S=1\text{S}=1.
    Mlinganyo sasa umesawazishwa.

    Jibu: 2NaOH(aq)+H2SO4(aq)→Na2SO4(aq)+2H2O(l)\boxed{2\text{NaOH(aq)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{Na}_2\text{SO}_4\text{(aq)} + 2\text{H}_2\text{O(l)}}

Part d· 3 steps

CuCO3(aq)+HNO3(aq)→Cu(NO3)2(aq)+CO2(g)+H2O(l)\text{CuCO}_3\text{(aq)} + \text{HNO}_3\text{(aq)} \rightarrow \text{Cu(NO}_3)_2\text{(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}

  1. Angalia idadi ya atomi za kila elementi pande zote mbili.

    Upande wa kushoto: Cu=1\text{Cu}=1, C=1\text{C}=1, O=3+3=6\text{O}=3+3=6, H=1\text{H}=1, N=1\text{N}=1.
    Upande wa kulia: Cu=1\text{Cu}=1, C=1\text{C}=1, O=2×3+2+1=9\text{O}=2 \times 3 + 2 + 1 = 9, H=2\text{H}=2, N=2\text{N}=2.

  2. Sawazisha atomi za nitrojeni (N\text{N}).

    Kuna atomi 2 za N\text{N} upande wa kulia na 1 upande wa kushoto. Weka koefishienti 2 mbele ya HNO3\text{HNO}_3.

    CuCO3(aq)+2HNO3(aq)→Cu(NO3)2(aq)+CO2(g)+H2O(l)\text{CuCO}_3\text{(aq)} + 2\text{HNO}_3\text{(aq)} \rightarrow \text{Cu(NO}_3)_2\text{(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}
  3. Angalia tena idadi ya atomi za kila elementi.

    Upande wa kushoto: Cu=1\text{Cu}=1, C=1\text{C}=1, O=3+(2×3)=9\text{O}=3+(2 \times 3)=9, H=2\text{H}=2, N=2\text{N}=2.
    Upande wa kulia: Cu=1\text{Cu}=1, C=1\text{C}=1, O=(2×3)+2+1=9\text{O}=(2 \times 3)+2+1=9, H=2\text{H}=2, N=2\text{N}=2.
    Mlinganyo sasa umesawazishwa.

    Jibu: CuCO3(aq)+2HNO3(aq)→Cu(NO3)2(aq)+CO2(g)+H2O(l)\boxed{\text{CuCO}_3\text{(aq)} + 2\text{HNO}_3\text{(aq)} \rightarrow \text{Cu(NO}_3)_2\text{(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}}

Part e· 4 steps

H2S(g)+O2(g)→SO2(g)+H2O(l)\text{H}_2\text{S(g)} + \text{O}_2\text{(g)} \rightarrow \text{SO}_2\text{(g)} + \text{H}_2\text{O(l)}

  1. Angalia idadi ya atomi za kila elementi pande zote mbili.

    Upande wa kushoto: H=2\text{H}=2, S=1\text{S}=1, O=2\text{O}=2.
    Upande wa kulia: H=2\text{H}=2, S=1\text{S}=1, O=2+1=3\text{O}=2+1=3.

  2. Sawazisha atomi za oksijeni (O\text{O}).

    Kuna atomi 2 za O\text{O} upande wa kushoto na 3 upande wa kulia. Ili kusawazisha, tutahitaji koefishienti zisizo kamili au kuzidisha mlinganyo mzima baadaye.
    Weka koefishienti 2 mbele ya H2S\text{H}_2\text{S} na 2 mbele ya SO2\text{SO}_2 na 2 mbele ya H2O\text{H}_2\text{O}.

    2H2S(g)+O2(g)→2SO2(g)+2H2O(l)2\text{H}_2\text{S(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{SO}_2\text{(g)} + 2\text{H}_2\text{O(l)}
  3. Sawazisha atomi za oksijeni (O\text{O}).

    Sasa kuna 2×2=42 \times 2 = 4 atomi za H\text{H} upande wa kushoto na 2×2=42 \times 2 = 4 upande wa kulia.
    Kuna 2×1=22 \times 1 = 2 atomi za S\text{S} upande wa kushoto na 2×1=22 \times 1 = 2 upande wa kulia.
    Kuna 2 atomi za O\text{O} upande wa kushoto na (2×2)+2=6(2 \times 2) + 2 = 6 upande wa kulia.
    Weka koefishienti 3 mbele ya O2\text{O}_2.

    2H2S(g)+3O2(g)→2SO2(g)+2H2O(l)2\text{H}_2\text{S(g)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{SO}_2\text{(g)} + 2\text{H}_2\text{O(l)}
  4. Angalia tena idadi ya atomi za kila elementi.

    Upande wa kushoto: H=4\text{H}=4, S=2\text{S}=2, O=3×2=6\text{O}=3 \times 2 = 6.
    Upande wa kulia: H=2×2=4\text{H}=2 \times 2 = 4, S=2\text{S}=2, O=(2×2)+2=6\text{O}=(2 \times 2)+2=6.
    Mlinganyo sasa umesawazishwa.

    Jibu: 2H2S(g)+3O2(g)→2SO2(g)+2H2O(l)\boxed{2\text{H}_2\text{S(g)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{SO}_2\text{(g)} + 2\text{H}_2\text{O(l)}}

Part f· 6 steps

C2H6(g)+O2(g)→CO2(g)+H2O(l)\text{C}_2\text{H}_6\text{(g)} + \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}

  1. Angalia idadi ya atomi za kila elementi pande zote mbili.

    Upande wa kushoto: C=2\text{C}=2, H=6\text{H}=6, O=2\text{O}=2.
    Upande wa kulia: C=1\text{C}=1, H=2\text{H}=2, O=2+1=3\text{O}=2+1=3.

  2. Sawazisha atomi za kaboni (C\text{C}).

    Kuna atomi 2 za C\text{C} upande wa kushoto na 1 upande wa kulia. Weka koefishienti 2 mbele ya CO2\text{CO}_2.

    C2H6(g)+O2(g)→2CO2(g)+H2O(l)\text{C}_2\text{H}_6\text{(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}
  3. Sawazisha atomi za hidrojeni (H\text{H}).

    Kuna atomi 6 za H\text{H} upande wa kushoto na 2 upande wa kulia. Weka koefishienti 3 mbele ya H2O\text{H}_2\text{O}.

    C2H6(g)+O2(g)→2CO2(g)+3H2O(l)\text{C}_2\text{H}_6\text{(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} + 3\text{H}_2\text{O(l)}
  4. Sawazisha atomi za oksijeni (O\text{O}).

    Sasa kuna 2 atomi za O\text{O} upande wa kushoto na (2×2)+(3×1)=4+3=7(2 \times 2) + (3 \times 1) = 4+3=7 upande wa kulia.
    Ili kusawazisha, weka koefishienti 72\frac{7}{2} mbele ya O2\text{O}_2.

    C2H6(g)+72O2(g)→2CO2(g)+3H2O(l)\text{C}_2\text{H}_6\text{(g)} + \frac{7}{2}\text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} + 3\text{H}_2\text{O(l)}
  5. Ondoa sehemu kwa kuzidisha mlinganyo mzima kwa 2.

    2C2H6(g)+7O2(g)→4CO2(g)+6H2O(l)2\text{C}_2\text{H}_6\text{(g)} + 7\text{O}_2\text{(g)} \rightarrow 4\text{CO}_2\text{(g)} + 6\text{H}_2\text{O(l)}
  6. Angalia tena idadi ya atomi za kila elementi.

    Upande wa kushoto: C=2×2=4\text{C}=2 \times 2 = 4, H=2×6=12\text{H}=2 \times 6 = 12, O=7×2=14\text{O}=7 \times 2 = 14.
    Upande wa kulia: C=4\text{C}=4, H=6×2=12\text{H}=6 \times 2 = 12, O=(4×2)+(6×1)=8+6=14\text{O}=(4 \times 2)+(6 \times 1)=8+6=14.
    Mlinganyo sasa umesawazishwa.

    Jibu: 2C2H6(g)+7O2(g)→4CO2(g)+6H2O(l)\boxed{2\text{C}_2\text{H}_6\text{(g)} + 7\text{O}_2\text{(g)} \rightarrow 4\text{CO}_2\text{(g)} + 6\text{H}_2\text{O(l)}}

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