Answer:
The maximum mass of water produced is [tex]m__{H_2O}} =116 \ g[/tex]
Explanation:
From the question we are told that
The mass of sucrose is [tex]m_s = 200 \ g[/tex]
The chemical formula for sucrose is [tex]C_{12} H_{22} O_{11}[/tex]
The chemical equation for the dissociation of sucrose is
[tex]C_{12} H_{22}O_{4} \to 12C + 11H_2O[/tex]
The number of moles of sucrose can be evaluated as
[tex]n = \frac{m}{Z}[/tex]
Where Z is the molar mass of sucrose which has a constant value of
[tex]Z = 342 \ g/mol[/tex]
So
[tex]n = \frac{200}{342}[/tex]
[tex]n =0.585[/tex]
From the chemical equation one mole of sucrose produces 11 moles of water so 0.585 moles of sucrose will produce x moles of water
Therefore
[tex]x = \frac{0.585 * 11}{1}[/tex]
[tex]x = 6.433 \ moles[/tex]
Now the mass of water produced is mathematically represented as
[tex]m__{H_2O}} = x * Z__{H_2O}[/tex]
Where [tex]Z__{H_2O}[/tex] is the molar mass of water with a constant values of [tex]Z__{H_2O}} = 18 \ g/mol[/tex]
So
[tex]m__{H_2O}} = 6.43* 18[/tex]
[tex]m__{H_2O}} =116 \ g[/tex]