(b) Given the equation CH_{4 (g)} + 2O_{2 (g)} ——→ CO_{2 (g)} + 2H_{2}O_{ (g)} + Heat, how can this be read in terms of particles, moles and masses?

**Solution**

The given equation is CH4_{ (g)} + 2O_{2 (g)} ——→ CO_{2 (g)} + 2H_{2}O_{ (g)} + Heat.

This equation can be read in a number of ways;

**In terms of particles**

1 molecule of CH_{4} + 2 molecules of O_{2} ———-→ 1 molecule of CO_{2 }+ 2 molecules of H_{2}O

Alternatively,

6.022 × 10^{23} molecules of CH_{4} + 2 × 6.022 × 10^{23} molecules of O_{2} ———-→ 6.022 × 10^{23} molecules of CO_{2} + 2 × 6.022 × 10^{23}molecules of H_{2}O

**In terms of moles**

Since 6.022 × 10^{23} molecules of CH_{4}, Or O_{2} or CO_{2} is one mole, therefore,

1 mol of CH_{4} + 2 mol of O_{2} ———–→ 1 mol of CO_{2} + 2 mol of H_{2}O

**In terms of masses**

Now one mol of CH_{4} is 16 g, 2 moles of O_{2} is 64 g, 1 mol of CO_{2} is 44 g and 2 moles of H_{2}O is 36 g, therefore,

16 g of CH_{4} + 64 g of O_{2} ———-→ 44 g of CO_{2} + 36 g of H_{2}O

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