Answer: 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
