Calculate the mass of H2OH2O produced by metabolism of 1.6 kgkg of fat, assuming the fat consists entirely of tristearin (C57H110O6C57H110O6), a typical animal fat, and assuming that during metabolism, tristearin reacts with O2O2 to form only CO2CO2 and H2OH2O.

1779.8g of H2O

Explanation:

C57H110O6 + O2 —> CO2 + H20

To calculate the amount of water(H2O) produced by the reaction of 1.6Kg tristearin(C57H110O6) and O2, we must first balance the equation.

To do so, add 2 in front of C57H110O6, 163 in front of O2, 114 in front of CO2 and 110 in front of H2O i.e

2C57H110O6 + 163O2 —> 114CO2 + 110H20

Next, we have to calculate the mass of tristearin(C57H110O6) that reacted and the mass of water produced from the balanced equation. This can be done by:

MM of C57H110O6 = (57x12) + (110x1) + (6x16) = 684 + 110 + 96 = 890g/mol

Mass conc of C57H110O6 = 890 x 2 = 1780g

MM of H2O = (2x1) + 16 = 18g/mol

Mass conc of H2O = 110 x 18 = 1980g

Next, we will find the amount of water produced by 1.6Kg(i.e 1600g) of tristearin(C57H110O6). This is done by:

From the equation,

1780g of C57H110O6 produced 1980g of H2O.

Therefore 1600g of C57H110O6 will produce = (1600x1980)/1780 = 1779.8g of H2O

im not exactly sure but i think its b!