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When the alkali metal cesium reacts with water, it forms cesium hydroxide and hydrogen gas. Us what you've learned about balancing equations to select the correct coefficients to represent the law of conservation of mass for the reaction below.

cesium + water → cesium hydroxide + hydrogen

Cs (s) + H2O (l) → CsOH (aq) +H2 (g)


(4 points)

Cs (s) +

H2O (l)
→

CsOH (aq) +

H2 (g)

1 answer

To balance the reaction of cesium with water, we first need to write down the unbalanced equation:

\[ \text{Cs (s)} + \text{H}_2\text{O (l)} \rightarrow \text{CsOH (aq)} + \text{H}_2\text{(g)} \]

Now, let's break it down:

  1. On the reactant side:

    • Cesium (Cs): 1
    • Water (H₂O): 2 hydrogen and 1 oxygen

  2. On the product side:

    • Cesium Hydroxide (CsOH): 1 cesium, 1 hydroxide contributes 1 oxygen and 1 hydrogen.
    • Hydrogen gas (H₂): 2 hydrogen

Balancing the equation:

  • There is 1 cesium on both sides, which is balanced.
  • There are 2 hydrogen atoms in total on the product side (1 from CsOH and 2 from H₂), and there are 2 hydrogen atoms in water on the reactant side. Therefore, this is also balanced.
  • There is 1 oxygen on both sides.

So the balanced equation is:

\[ \text{Cs (s)} + \text{H}_2\text{O (l)} \rightarrow \text{CsOH (aq)} + \text{H}_2\text{(g)} \]

The coefficients are:

  • 1 for Cs
  • 1 for H₂O
  • 1 for CsOH
  • 1 for H₂

Thus, the final balanced equation is:

1 Cs (s) + 1 H₂O (l) → 1 CsOH (aq) + 1 H₂ (g)

In summary, you do not need to change any coefficients from the original single entries. It is already correctly balanced with each being 1.