Question

For the following reaction, the mass of water produced from $445\mathrm{g}$ of ${\mathrm{C}}_{57}{\mathrm{H}}_{110}{\mathrm{O}}_6$ is:

$\underset{\mathrm{Stearin}}{\underbrace{2{\mathrm{C}}_{57}{\mathrm{H}}_{110}{\mathrm{O}}_6\left(\mathrm{s}\right)}}+\underset{\mathrm{Oxygen}}{\underbrace{163{\mathrm{O}}_2\left(\mathrm{g}\right)}}\to \underset{\mathrm{Carbondioxide}}{\underbrace{114{\mathrm{CO}}_2\left(\mathrm{g}\right)}}+\underset{\mathrm{water}}{\underbrace{110{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}}$

Answer 1

Step 1: Given data

Mass of ${\mathrm{C}}_{57}{\mathrm{H}}_{110}{\mathrm{O}}_6$ is $445\mathrm{g}$

Step 2: Calculating the Mass of water

Molecular mass of ${\mathrm{C}}_{57}{\mathrm{H}}_{110}{\mathrm{O}}_6=890\mathrm{g}$

Mole of ${\mathrm{C}}_{57}{\mathrm{H}}_{110}{\mathrm{O}}_6$ is $(\mathrm{weight}/\mathrm{molecular}\mathrm{weight})$$=\frac{445}{890}=0.5\mathrm{mole}$

From the equation, we can see that $2\mathrm{mole}{\mathrm{C}}_{57}{\mathrm{H}}_{110}{\mathrm{O}}_6$ produces $110\mathrm{moles}$of ${\mathrm{H}}_2\mathrm{O}$

Apply stoichiometry,

$0.5\mathrm{mole}$produces $\frac{110}{2}\times 0.5=27.5$

The molecular weight of ${\mathrm{H}}_2\mathrm{O}$ is $18$

The mass of water $=27.5\times 18=495\mathrm{gm}$

Therefore, mass of water produced is $495\mathrm{gm}$.