High School

How many milliliters of [tex]C_5H_{12}[/tex] are needed to make 97.3 mL of [tex]C_5H_8[/tex]?

Density of [tex]C_5H_{12}[/tex] = 0.620 g/mL
Density of [tex]C_5H_8[/tex] = 0.681 g/mL
Density of [tex]H_2[/tex] = 0.0899 g/L

Please explain.

Answer :

First of all, we can write a chemical equation.

[tex]C _{5}H_{12} = C _{5}H_{8} + 2H_{2} [/tex]

Thus, we must use 1 mole C₅H₁₂ to obtain 1 mole C₅H₈ and 2 moles of hydrogen gas. According to conservation of mass law, products and reactants must have the same total mass.

97.3 ml C₅H₈= (97.3).(0.681)=66.26 gr. Moreover we can calculate its mole with molecular weight; C₅H₈=68 g/mole
Mole of C₅H₈=66.26/68=0.97
Mole of H₂= 1.94 Mass=1,94.2=3.89g Volume=3.89/0.0899=43.35 ml
Mole of C₅H₁₂=0.97 Mass=69.84 Volume=69.84/0.620=112.64 ml