Get reliable answers to your questions at Westonci.ca, where our knowledgeable community is always ready to help. Get quick and reliable solutions to your questions from a community of experienced professionals on our platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.

How many milliliters of C5H12 are needed to make 97.3 ml C5H8?

density of C5H12 = 0.620 g/mL
density of C5H8 = 0.681 g/mL
density of H2= 0.0899g/L

Please explain


Sagot :

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