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)

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Asked on June 27, 2026Chemistry

This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.

Accepted Answer · 2026-06-27T07:07:32.053Z

Haya njirudionisia69, tuendelee na maswali ya kusawazisha milinganyo. 12. Sawazisha milinganyo ifuatayo: b) Na(s) + H_2O(l) → NaOH(aq) + H_2(g) Step 1: Angalia idadi ya atomi za kila elementi pande zote mbili. Upande wa kushoto: Na=1, H=2, O=1. Upande wa kulia: Na=1, H=1+2=3, O=1. Step 2: Sawazisha atomi za hidrojeni na oksijeni. Weka koefishienti 2 mbele ya H_2O na 2 mbele ya NaOH na 2 mbele ya Na. 2Na(s) + 2H_2O(l) → 2NaOH(aq) + H_2(g) Step 3: Angalia tena idadi ya atomi za kila elementi. Upande wa kushoto: Na=2, H=2 × 2 = 4, O=2. Upande wa kulia: Na=2, H=2+2=4, O=2. Mlinganyo sasa umesawazishwa. Jibu: 2Na(s) + 2H_2O(l) → 2NaOH(aq) + H_2(g) c) NaOH(aq) + H_2SO_4(aq) → Na_2SO_4(aq) + H_2O(l) Step 1: Angalia idadi ya atomi za kila elementi pande zote mbili. Upande wa kushoto: Na=1, O=1+4=5, H=1+2=3, S=1. Upande wa kulia: Na=2, O=4+1=5, H=2, S=1. Step 2: Sawazisha atomi za sodiamu (Na). Kuna atomi 2 za Na upande wa kulia na 1 upande wa kushoto. Weka koefishienti 2 mbele ya NaOH. 2NaOH(aq) + H_2SO_4(aq) → Na_2SO_4(aq) + H_2O(l) Step 3: Sawazisha atomi za hidrojeni (H) na oksijeni (O). Sasa kuna 2 × 1 + 2 = 4 atomi za H upande wa kushoto na 2 upande wa kulia. Weka koefishienti 2 mbele ya H_2O. 2NaOH(aq) + H_2SO_4(aq) → Na_2SO_4(aq) + 2H_2O(l) Step 4: Angalia tena idadi ya atomi za kila elementi. Upande wa kushoto: Na=2, O=2+4=6, H=2+2=4, S=1. Upande wa kulia: Na=2, O=4+2=6, H=2 × 2 = 4, S=1. Mlinganyo sasa umesawazishwa. Jibu: 2NaOH(aq) + H_2SO_4(aq) → Na_2SO_4(aq) + 2H_2O(l) d) CuCO_3(aq) + HNO_3(aq) → Cu(NO_3)_2(aq) + CO_2(g) + H_2O(l) Step 1: Angalia idadi ya atomi za kila elementi pande zote mbili. Upande wa kushoto: Cu=1, C=1, O=3+3=6, H=1, N=1. Upande wa kulia: Cu=1, C=1, O=2 × 3 + 2 + 1 = 9, H=2, N=2. Step 2: Sawazisha atomi za nitrojeni (N). Kuna atomi 2 za N upande wa kulia na 1 upande wa kushoto. Weka koefishienti 2 mbele ya HNO_3. CuCO_3(aq) + 2HNO_3(aq) → Cu(NO_3)_2(aq) + CO_2(g) + H_2O(l) Step 3: Angalia tena idadi ya atomi za kila elementi. Upande wa kushoto: Cu=1, C=1, O=3+(2 × 3)=9, H=2, N=2. Upande wa kulia: Cu=1, C=1, O=(2 × 3)+2+1=9, H=2, N=2. Mlinganyo sasa umesawazishwa. Jibu: CuCO_3(aq) + 2HNO_3(aq) → Cu(NO_3)_2(aq) + CO_2(g) + H_2O(l) e) H_2S(g) + O_2(g) → SO_2(g) + H_2O(l) Step 1: Angalia idadi ya atomi za kila elementi pande zote mbili. Upande wa kushoto: H=2, S=1, O=2. Upande wa kulia: H=2, S=1, O=2+1=3. Step 2: Sawazisha atomi za oksijeni (O). Kuna atomi 2 za 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 H_2S na 2 mbele ya SO_2 na 2 mbele ya H_2O. 2H_2S(g) + O_2(g) → 2SO_2(g) + 2H_2O(l) Step 3: Sawazisha atomi za oksijeni (O). Sasa kuna 2 × 2 = 4 atomi za H upande wa kushoto na 2 × 2 = 4 upande wa kulia. Kuna 2 × 1 = 2 atomi za S upande wa kushoto na 2 × 1 = 2 upande wa kulia. Kuna 2 atomi za O upande wa kushoto na (2 × 2) + 2 = 6 upande wa kulia. Weka koefishienti 3 mbele ya O_2. 2H_2S(g) + 3O_2(g) → 2SO_2(g) + 2H_2O(l) Step 4: Angalia tena idadi ya atomi za kila elementi. Upande wa kushoto: H=4, S=2, O=3 × 2 = 6. Upande wa kulia: H=2 × 2 = 4, S=2, O=(2 × 2)+2=6. Mlinganyo sasa umesawazishwa. Jibu: 2H_2S(g) + 3O_2(g) → 2SO_2(g) + 2H_2O(l) f) C_2H_6(g) + O_2(g) → CO_2(g) + H_2O(l) Step 1: Angalia idadi ya atomi za kila elementi pande zote mbili. Upande wa kushoto: C=2, H=6, O=2. Upande wa kulia: C=1, H=2, O=2+1=3. Step 2: Sawazisha atomi za kaboni (C). Kuna atomi 2 za C upande wa kushoto na 1 upande wa kulia. Weka koefishienti 2 mbele ya CO_2. C_2H_6(g) + O_2(g) → 2CO_2(g) + H_2O(l) Step 3: Sawazisha atomi za hidrojeni (H). Kuna atomi 6 za H upande wa kushoto na 2 upande wa kulia. Weka koefishienti 3 mbele ya H_2O. C_2H_6(g) + O_2(g) → 2CO_2(g) + 3H_2O(l) Step 4: Sawazisha atomi za oksijeni (O). Sasa kuna 2 atomi za O upande wa kushoto na (2 × 2) + (3 × 1) = 4+3=7 upande wa kulia. Ili kusawazisha, weka koefishienti (7)/(2) mbele ya O_2. C_2H_6(g) + (7)/(2)O_2(g) → 2CO_2(g) + 3H_2O(l) Step 5: Ondoa sehemu kwa kuzidisha mlinganyo mzima kwa 2. 2C_2H_6(g) + 7O_2(g) → 4CO_2(g) + 6H_2O(l) Step 6: Angalia tena idadi ya atomi za kila elementi. Upande wa kushoto: C=2 × 2 = 4, H=2 × 6 = 12, O=7 × 2 = 14. Upande wa kulia: C=4, H=6 × 2 = 12, O=(4 × 2)+(6 × 1)=8+6=14. Mlinganyo sasa umesawazishwa. Jibu: 2C_2H_6(g) + 7O_2(g) → 4CO_2(g) + 6H_2O(l) That's all for this page! Send me the next one 📸

Accepted Answer · 2026-06-27T07:07:32.053Z

Haya njirudionisia69, tuendelee na maswali ya kusawazisha milinganyo. 12. Sawazisha milinganyo ifuatayo: b) Na(s) + H_2O(l) → NaOH(aq) + H_2(g) Step 1: Angalia idadi ya atomi za kila elementi pande zote mbili. Upande wa kushoto: Na=1, H=2, O=1. Upande wa kulia: Na=1, H=1+2=3, O=1. Step 2: Sawazisha atomi za hidrojeni na oksijeni. Weka koefishienti 2 mbele ya H_2O na 2 mbele ya NaOH na 2 mbele ya Na. 2Na(s) + 2H_2O(l) → 2NaOH(aq) + H_2(g) Step 3: Angalia tena idadi ya atomi za kila elementi. Upande wa kushoto: Na=2, H=2 × 2 = 4, O=2. Upande wa kulia: Na=2, H=2+2=4, O=2. Mlinganyo sasa umesawazishwa. Jibu: 2Na(s) + 2H_2O(l) → 2NaOH(aq) + H_2(g) c) NaOH(aq) + H_2SO_4(aq) → Na_2SO_4(aq) + H_2O(l) Step 1: Angalia idadi ya atomi za kila elementi pande zote mbili. Upande wa kushoto: Na=1, O=1+4=5, H=1+2=3, S=1. Upande wa kulia: Na=2, O=4+1=5, H=2, S=1. Step 2: Sawazisha atomi za sodiamu (Na). Kuna atomi 2 za Na upande wa kulia na 1 upande wa kushoto. Weka koefishienti 2 mbele ya NaOH. 2NaOH(aq) + H_2SO_4(aq) → Na_2SO_4(aq) + H_2O(l) Step 3: Sawazisha atomi za hidrojeni (H) na oksijeni (O). Sasa kuna 2 × 1 + 2 = 4 atomi za H upande wa kushoto na 2 upande wa kulia. Weka koefishienti 2 mbele ya H_2O. 2NaOH(aq) + H_2SO_4(aq) → Na_2SO_4(aq) + 2H_2O(l) Step 4: Angalia tena idadi ya atomi za kila elementi. Upande wa kushoto: Na=2, O=2+4=6, H=2+2=4, S=1. Upande wa kulia: Na=2, O=4+2=6, H=2 × 2 = 4, S=1. Mlinganyo sasa umesawazishwa. Jibu: 2NaOH(aq) + H_2SO_4(aq) → Na_2SO_4(aq) + 2H_2O(l) d) CuCO_3(aq) + HNO_3(aq) → Cu(NO_3)_2(aq) + CO_2(g) + H_2O(l) Step 1: Angalia idadi ya atomi za kila elementi pande zote mbili. Upande wa kushoto: Cu=1, C=1, O=3+3=6, H=1, N=1. Upande wa kulia: Cu=1, C=1, O=2 × 3 + 2 + 1 = 9, H=2, N=2. Step 2: Sawazisha atomi za nitrojeni (N). Kuna atomi 2 za N upande wa kulia na 1 upande wa kushoto. Weka koefishienti 2 mbele ya HNO_3. CuCO_3(aq) + 2HNO_3(aq) → Cu(NO_3)_2(aq) + CO_2(g) + H_2O(l) Step 3: Angalia tena idadi ya atomi za kila elementi. Upande wa kushoto: Cu=1, C=1, O=3+(2 × 3)=9, H=2, N=2. Upande wa kulia: Cu=1, C=1, O=(2 × 3)+2+1=9, H=2, N=2. Mlinganyo sasa umesawazishwa. Jibu: CuCO_3(aq) + 2HNO_3(aq) → Cu(NO_3)_2(aq) + CO_2(g) + H_2O(l) e) H_2S(g) + O_2(g) → SO_2(g) + H_2O(l) Step 1: Angalia idadi ya atomi za kila elementi pande zote mbili. Upande wa kushoto: H=2, S=1, O=2. Upande wa kulia: H=2, S=1, O=2+1=3. Step 2: Sawazisha atomi za oksijeni (O). Kuna atomi 2 za 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 H_2S na 2 mbele ya SO_2 na 2 mbele ya H_2O. 2H_2S(g) + O_2(g) → 2SO_2(g) + 2H_2O(l) Step 3: Sawazisha atomi za oksijeni (O). Sasa kuna 2 × 2 = 4 atomi za H upande wa kushoto na 2 × 2 = 4 upande wa kulia. Kuna 2 × 1 = 2 atomi za S upande wa kushoto na 2 × 1 = 2 upande wa kulia. Kuna 2 atomi za O upande wa kushoto na (2 × 2) + 2 = 6 upande wa kulia. Weka koefishienti 3 mbele ya O_2. 2H_2S(g) + 3O_2(g) → 2SO_2(g) + 2H_2O(l) Step 4: Angalia tena idadi ya atomi za kila elementi. Upande wa kushoto: H=4, S=2, O=3 × 2 = 6. Upande wa kulia: H=2 × 2 = 4, S=2, O=(2 × 2)+2=6. Mlinganyo sasa umesawazishwa. Jibu: 2H_2S(g) + 3O_2(g) → 2SO_2(g) + 2H_2O(l) f) C_2H_6(g) + O_2(g) → CO_2(g) + H_2O(l) Step 1: Angalia idadi ya atomi za kila elementi pande zote mbili. Upande wa kushoto: C=2, H=6, O=2. Upande wa kulia: C=1, H=2, O=2+1=3. Step 2: Sawazisha atomi za kaboni (C). Kuna atomi 2 za C upande wa kushoto na 1 upande wa kulia. Weka koefishienti 2 mbele ya CO_2. C_2H_6(g) + O_2(g) → 2CO_2(g) + H_2O(l) Step 3: Sawazisha atomi za hidrojeni (H). Kuna atomi 6 za H upande wa kushoto na 2 upande wa kulia. Weka koefishienti 3 mbele ya H_2O. C_2H_6(g) + O_2(g) → 2CO_2(g) + 3H_2O(l) Step 4: Sawazisha atomi za oksijeni (O). Sasa kuna 2 atomi za O upande wa kushoto na (2 × 2) + (3 × 1) = 4+3=7 upande wa kulia. Ili kusawazisha, weka koefishienti (7)/(2) mbele ya O_2. C_2H_6(g) + (7)/(2)O_2(g) → 2CO_2(g) + 3H_2O(l) Step 5: Ondoa sehemu kwa kuzidisha mlinganyo mzima kwa 2. 2C_2H_6(g) + 7O_2(g) → 4CO_2(g) + 6H_2O(l) Step 6: Angalia tena idadi ya atomi za kila elementi. Upande wa kushoto: C=2 × 2 = 4, H=2 × 6 = 12, O=7 × 2 = 14. Upande wa kulia: C=4, H=6 × 2 = 12, O=(4 × 2)+(6 × 1)=8+6=14. Mlinganyo sasa umesawazishwa. Jibu: 2C_2H_6(g) + 7O_2(g) → 4CO_2(g) + 6H_2O(l) That's all for this page! Send me the next one 📸

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)

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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)

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