Quiz | Chemical Calculations

n_benzaldehyde  =  $\frac{{\mathrm{m}}_{\mathrm{benzaldehyde}}}{{\mathrm{M}}_{\mathrm{benzaldehyde}}}$   ⇒   m_benzaldehyde = n_benzaldehyde x M_benzaldehyde 

M_benzaldehyde = 7 x M_C + 6 x M_H + M_O = 7 x 12.0 + 6 x 1.0 + 16.0 = 106.0 g.mol^-1

⇒ m_benzaldehyde = 1 x 106.0 = 106 g

Carbon dioxide formula: CO₂. There is 1 mole of carbon in one mole of CO₂

M_C = 12.0 g.mol^-1;   M_CO2 = M_C + 2 x M_O = 12.0 + 2 x 16.0 = 44.0 g.mol^-1

%mass of C in CO₂ = moles of C in CO₂ x $\frac{{\mathrm{M}}_{\mathrm{C}}}{{\mathrm{M}}_{\mathrm{CO}2}}$ x 100  =  1 x $\frac{12.0}{44.0}$ x 100 = 27.3 %

mass of C in CO₂ = $\frac{\%\mathrm{mass} \mathrm{of} \mathrm{C} \mathrm{in} {\mathrm{CO}}_2}{100}$ x mass of CO₂ = 0.273 x 110 = 30.0 g

n_H = $\frac{{\mathrm{m}}_{\mathrm{H}}}{{\mathrm{M}}_{\mathrm{H}}}$ = $\frac{10}{1.01}$ = 9.90 moles

There are 2 moles of H in 1 mole of H₂   ⇒   n_H2 = $\raisebox{1ex}{${\mathrm{n}}_{\mathrm{H}}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ = 4.95 moles

n_C  =  $\frac{\mathrm{N}}{{\mathrm{N}}_{\mathrm{A}}}$ ⇒ Number of carbon atoms N = n_C x N_A 

There are 2 moles of C in 1 mole of C₂H₄: n_C = 2 x n_C2H4 

⇒ N = n_C x N_A = 2 x n_C2H4 x N_A = 2 x 0.4 x 6.022 x 10²³ = 4.82 x 10²³ atoms

The limiting reagent is the reagent completely consumed which determines the quantity of products formed

n_N2 = $\frac{{\mathrm{m}}_{\mathrm{N}2}}{{\mathrm{M}}_{\mathrm{N}2}}$= $\frac{5.0}{28.0}$ = 0.179 mole;   n_H2 = $\frac{{\mathrm{m}}_{\mathrm{H}2}}{{\mathrm{M}}_{\mathrm{H}2}}$ = $\frac{5.0}{2.02}$ = 2.47 moles

Chemical Equation: 3 H₂ + N₂ → 2 NH₃ 
⇒ 3 molecules of H₂ react with 1 molecule of N₂ to form 2 molecules of NH₃

n_N2 < $\frac{{\mathrm{n}}_{\mathrm{H}2}}{3}$ ⇒ N₂ is the limiting reagent and n_NH3 = 2 x n_N2 = 0.357 mole

m_NH3 = n_NH3 x M_NH3 = 0.357 x 17.0 = 6.08 g