Question

part b – balancing practice (show your work)
balance the following equations. remember: atoms are never lost or gained, only rearranged.

1. h₂ + o₂ → h₂o
2. na + cl₂ → nacl
3. mg + o₂ → mgo

Explanation:

1. Balancing $\boldsymbol{\ce{H_2 + O_2

ightarrow H_2O}}$

Step1: Count O atoms

Left: 2 O (from $\ce{O_2}$), Right: 1 O (from $\ce{H_2O}$). Multiply $\ce{H_2O}$ by 2: $\ce{H_2 + O_2
ightarrow 2H_2O}$

Step2: Count H atoms

Left: 2 H (from $\ce{H_2}$), Right: 4 H (from $2\ce{H_2O}$). Multiply $\ce{H_2}$ by 2: $\ce{2H_2 + O_2
ightarrow 2H_2O}$

Step1: Count Cl atoms

Left: 2 Cl (from $\ce{Cl_2}$), Right: 1 Cl (from $\ce{NaCl}$). Multiply $\ce{NaCl}$ by 2: $\ce{Na + Cl_2
ightarrow 2NaCl}$

Step2: Count Na atoms

Left: 1 Na, Right: 2 Na (from $2\ce{NaCl}$). Multiply $\ce{Na}$ by 2: $\ce{2Na + Cl_2
ightarrow 2NaCl}$

Step1: Count O atoms

Left: 2 O (from $\ce{O_2}$), Right: 1 O (from $\ce{MgO}$). Multiply $\ce{MgO}$ by 2: $\ce{Mg + O_2
ightarrow 2MgO}$

Step2: Count Mg atoms

Left: 1 Mg, Right: 2 Mg (from $2\ce{MgO}$). Multiply $\ce{Mg}$ by 2: $\ce{2Mg + O_2
ightarrow 2MgO}$

Answer:

$\boldsymbol{\ce{2H_2 + O_2
ightarrow 2H_2O}}$

2. Balancing $\boldsymbol{\ce{Na + Cl_2

ightarrow NaCl}}$