Question

1a) 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?

1b) What volume of nitrogen dioxide is formed at 705 torr and 28.2 °C by reacting 1.95 cm3 of copper (d = 8.95 g/cm3) with 1.50 L of nitric acid 2.53 M?

Answer 1

1a) molar mass of C₁₈H₃₀O₂ = 278.43 gm/mol

molar mass of H₂ = 2.01 gm/mol then 3 mole of  H₂ = 6.03 gm

According to reaction 1 mole of C₁₈H₃₀O₂ react with 3 mole of H₂ that mean 278.43 gm of C₁₈H₃₀O₂ react with 6.03 gm of H₂ then to react with 10.5 gm C₁₈H₃₀O₂ require = 10.5 6.03 / 278.43 = 0.2274 gm of H₂

molar mass of H₂ = 2.01 gm/mol then 0.2274 gm = 0.2274/2.01 = 0.1131 mole of H₂ gas

at STP 1mole of gas occupy volume 22.414 L then 0.1131 mole gas occupy volume = 22.414 0.1134 = 2.54 L

2.54 L of H₂ require to react with 10.5 gm of linolenic acid at STP

1b) reaction is

Cu₂ + 8HNO₃    2Cu(NO₃)₂ + 4NO₂ + 4H₂O

mass = density volume

mass of Cu = 8.95 1.95 = 17.4525 gm

molar mass of Cu₂ = 127.098 gm/mol then 17.4525 gm = 17.4525/127.098 = 0.1373 mole of Cu

no. of mole = molarity volume of solution in liter

no. of mole of HNO₃  = 2.53 1.50 = 3.795 mole

According to reaction 1 mole of Cu₂ react with 8 mole of HNO₃ then to react with 0.1373 mole of Cu₂ require

0.1373 8 = 1.0985 mole of HNO₃ but HNO₃ given 3.795 mole thus HNO₃ is excess reactant and Cu is limiting reactant react completly

According to reaction 1 mole Cu₂ produce 4 mole of NO₂ then 0.1373 mole of Cu₂ produce 0.1373 4 = 0.5492 mole of NO₂

calculate volume of NO₂ by using ideal gas equation

We know that PV = nRT

V = nRT/P

n = 0.5492 mole,

T = 28.2⁰C = 28.2+273.15 = 301.35K,

P= 705 torr = 0.927632 atm,

R = 0.08205 L atm mol^-1 K^-1 ( R = gas constant)

V = ?

Substitute these value in above equation.

V = 0.5492 0.08205 301.35/0.927632 = 14.64 L

14.64 L NO₂ formed.