How many molecules of CO_(2)will needed to obtain 1.8 g of glucose according to given reaction. Reaction : 6CO_(2) + 6H_(2) O rarr C_(6) H_(12) O_(6) + 6O_(2) [M. wt C_(6)H_(12)O_(6) = 180 g mol^(-1)] (C=12, H=1, O=16)

How many molecules of CO_(2)will needed to obtain 1.8 g of glucose acc

1.	How many molecules of CO_(2)will needed to obtain 1.8 g of glucose according to given reaction. Reaction : 6CO_(2) + 6H_(2) O rarr C_(6) H_(12) O_(6) + 6O_(2) [M. wt C_(6)H_(12)O_(6) = 180 g mol^(-1)] (C=12, H=1, O=16)
Answer» `0.6xx6.022xx10^(23)` `6xx6.022xx10^(23)` `0.06xx6.022xx10^(23)` `60xx6.022xx10^(23)` Solution :mole of glucose `=("mass ")/("molecular mass") = (1.8)/(180)` ` =0.01 ` mole glucose ...(1) Reaction : `6CO_(2) + 6H_(2) O RARR C_(6)H_(12)O_(6) + 6O_(2)` `:.` 6 mole `CO_(2)` is needed to GET 1 mole glucose `= (0.01xx6)/(1)` `= 0.06` mole `CO_(2)` is needed ...(ii) `=0.06xx6.022xx10^(23)` molecular of `CO_(2)`. ...(iii)