Linolenic acid (C18H30O2) can be hydrogenated to stearic acid by reacting it with hydrogen gas according to the equation: C18H30O2 + 3H2 --->C18H36O2 What volume of hydrogen gas, measured at STP, is required to react with 10.5 g of linolenic acid in this reaction?
Answers
1.998 ml of hydrogen gas volume is is required to react with 10.5 g of linolenic acid in this reaction.
Explanation:
Data given:
mass of linolenic acid is given = 10.5 grams
atomic mass of linoleinic acid = 278.43 grams/mole
volume of hydrogen required at STP = ?
Balance chemical reaction:
C18H30O2 + 3H2 --->C18H36O2
moles of linoleinic acid given =
putting the values in the equation:
moles of linoleinic acid =
= 0.037 moles
from the balanced equation:
1 mole of linoleinic acid reacts with 3 moles of hydrogen
so, 0.037 moles will react with x moles of hydrogen gas
=
x = 0.111 moles of water
volume or mass = atomic mass x number of moles
volume of hydrogen gas= 1.998 ml
Final answer:
Around 2.53 liters of hydrogen gas, at STP, is required to fully hydrogenate 10.5 g linolenic acid to stearic acid.
Explanation:
To solve this problem, we first need to find the molar mass of linolenic acid (C18H30O2). The molar mass is calculated by adding the molar masses of all the atoms in the molecule: (18*12.01) + (30*1.01) + (2*16.00) = 278.44 g/mol. Next, we calculate the number of moles of the linolenic acid we have: 10.5 g / 278.44 g/mol = 0.0377 mol. According to the balanced chemical equation, we need 3 mol of hydrogen per mol of Linolenic acid, so we need 0.0377 mol * 3 = 0.1131 mol of hydrogen gas. Finally, at STP, 1 mole of any type of gas occupies a volume of roughly 22.4 L, so the volume required is 0.1131 mol * 22.4 L/mol = 2.53 L.
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