Question

A drug used to treat enlarged prostates undergoes the following theoretical reaction: C17H30F6N2O2(s) ------ > CO2(g) + 2 NH3(g) + 6 HF(g) + C8H18(s) + 8 C(s) (a) How many grams of HF(g) can be produced from 55.0 grams of C17H30F6N2O2(s)? (b) In a separate experiment, how many atoms of C(s) can be obtained from 0.0000725 moles of C17H30F6N2O2(s) ?

Answer 1

molar weight of C₁₇H₃₀F₆N₂O₂ = (12*17+1*30+19*6+14*2+16*2) =408 g/mol

now 55 g of C₁₇H₃₀F₆N₂O₂ =(55/408) = 0.135 moles C₁₇H₃₀F₆N₂O₂

from 1 mole of C₁₇H₃₀F₆N₂O₂ , HF(g) is being produced = 6moles= 6(1*1+19*1) = 120 g

so, from 0.135 moles of C₁₇H₃₀F₆N₂O₂ , HF(g) is being produced = (6*0.135) moles= 6(1*1+19*1)*0.135=16.2 g HF(g)

from 1 mole of C₁₇H₃₀F₆N₂O₂ , Carbon(s) is being prouced = 8 moles =(8* 6.023*10²³ ) atoms

from 0.0000725 moles of C₁₇H₃₀F₆N₂O₂ , Carbon(s) is being prouced = (8 *0.0000725) moles =(8* 6.023*10²³ *0.0000725) =3.49*10²⁰ atoms of carbon

hope it will help you.plz rate me