Question

Can someone explain - For the following reaction , the mass of water produced from 445 g of is : - Some basic concepts in chemistry - JEE Main
For the following reaction , the mass of water produced from 445 g of C_{57}H_{110}O_{6} is :

2C_{57}H_{110}O_{6}(s)+ 163O_{2}(g)\rightarrow 114CO_{2}(g)+110H_{2}O(1)

Answer 1

reaction is $2C_{57}H_{110}O_{6}(s)+ 163O_{2}(g)\rightarrow 114CO_{2}(g)+110H_{2}O$

here you can see that 2 mole of $C_{57}H_{110}O_{6}$ produced 110 moles of water.

so, 1 mole of $C_{57}H_{110}O_{6}$ produced 55 moles of water.

or, 890 g of $C_{57}H_{110}O_{6}$ produced 55 × 18 = 990g of water

[ molecular weight of $C_{57}H_{110}O_{6}$ = 57 × 12 + 110 + 6 × 16 = 890g/mol and molecular weight of water = 18g/mol ]

or, 445 g of $C_{57}H_{110}O_{6}$ produced 495g of water .

hence, the mass of water produced from 445 g of $C_{57}H_{110}O_{6}$ is 495g

Answer 2

The given reaction is:

Now the molar mass of the given compounds is as follows:

C57H110O6 = 890 gram

O2 = 32 grams

CO2 = 44 gram

H20 = 18 gram

From the reaction, we can conclude that 2 moles of C57H110O6 produces 110 moles of H20

2 moles of C57H110O6 = 1780 grams

110 moles of H2O = 1980 grams of water

hence, in other words we can write:

1780 grams of C57H110O6 produces 1980 grams of water

Hence, 445 g of C57H110O6 will produce (1980/1780) x 445 grams of water

On calculating, it is equal to 495 grams of water.