You have to start by balancing the equation
This is the equation balanced:
\(2 C_{57} H_{110} O_{6} +163 O_{2} -\ \textgreater \ 114CO_{2} +110 H_{2} O\)
Relation 2 mols of triestarin: 110 mol of water
To pass 2 mols of triesetarin you have to multiply by the molar mass of the compound.
The molar mass of the triestarin is calculated as 57atomic mass of C + 110 atomic mass of H + 6atomic mass of oxygen = 5712 + 1101 + 616 = 890 g/mol
Then 2 mols = 2 890 g/mol = 1780 g
Now pass 110 mol of water to grams using the molar mass of water (18 g/mol)
110 mol 18 g/mol = 1980 g of water
Now the relation in grams is 1780 g of tristearin / 1980 g of water = 3000 g of triestarin / x grams of water.
Solving for x, x = 3000 g of triestarin 1980 g of water /1780 g of triestarin
x = 3337 g of water = 3.3 kg of water
Answer: 3.3 kg of water
