Question

balancing equations

1. h₃po₄ + koh → k₃po₄ + h₂o
2. k + b₂o₃ → k₂o + b
3. hcl + naoh → nacl + h₂o
4. na + nano₃ → na₂o + n₂
5. c + s₈ → cs₂
6. na + o₂ → na₂o
7. n₂ + o₂ → n₂o₅
8. h₃po₄ + mg(oh)₂ → mg₃(po₄)₂ + h₂o
9. naoh + h₂co₃ → na₂co₃ + h₂o
10. koh + hbr → kbr + h₂o
11. na + o₂ → na₂o
12. al(oh)₃ + h₂co₃ → al₂(co₃)₃ + h₂o
13. al + s₈ → al₂s₃
14. cs + n₂ → cs₃n
15. mg + cl₂ → mgcl₂
16. rb + rbno₃ → rb₂o + n₂
17. c₆h₆ + o₂ → co₂ + h₂o
18. n₂ + h₂ → nh₃
19. c₁₀h₂₂ + o₂ → co₂ + h₂o
20. al(oh)₃ + hbr → albr₃ + h₂o
21. ch₃ch₂ch₂ch₃ + o₂ → co₂ + h₂o
22. c₃h₈ + o₂ → co₂ + h₂o
23. li + alcl₃ → licl + al
24. c₂h₆ + o₂ → co₂ + h₂o
25. nh₄oh + h₃po₄ → (nh₄)₃po₄ + h₂o
26. rb + p → rb₃p
27. ch₄ + o₂ → co₂ + h₂o
28. al(oh)₃ + h₂so₄ → al₂(so₄)₃ + h₂o
29. na + cl₂ → nacl
30. rb + s₈ → rb₂s
31. h₃po₄ + ca(oh)₂ → ca₃(po₄)₂ + h₂o
32. nh₃ + hcl → nh₄cl
33. li + h₂o → lioh + h₂
34. ca₃(po₄)₂ + sio₂ + c → casio₃ + co +
35. nh₃ + o₂ → n₂ + h₂o
36. fes₂ + o₂ → fe₂o₃ + so₂
37. c + so₂ → cs₂ + co

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Explanation:

Step1: Analyze the number of atoms for each element

For the equation \(N_2 + H_2
ightarrow NH_3\), on the left - hand side, there are \(2\) nitrogen atoms (\(N\)) in \(N_2\) and \(2\) hydrogen atoms (\(H\)) in \(H_2\). On the right - hand side, there is \(1\) nitrogen atom and \(3\) hydrogen atoms in \(NH_3\).

Step2: Balance the nitrogen atoms

To balance the nitrogen atoms, we put a coefficient of \(2\) in front of \(NH_3\). The equation becomes \(N_2 + H_2
ightarrow2NH_3\). Now, the number of nitrogen atoms is balanced (\(2\) on each side).

Step3: Balance the hydrogen atoms

After balancing nitrogen, for hydrogen, on the right - hand side, there are \(6\) hydrogen atoms (\(2\times3\) in \(2NH_3\)). So, we put a coefficient of \(3\) in front of \(H_2\). The balanced equation is \(N_2 + 3H_2 = 2NH_3\).

Answer:

\(1N_2+3H_2 = 2NH_3\)