Quiz 1 - Chemical Calculations | 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

Molarity M  =  $\frac{{\mathrm{n}}_{\mathrm{solute}}}{{\mathrm{V}}_{\mathrm{solution}}}$  =  $\frac{8}{2}$ =  4M

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

number of moles n_NaCl  =  M x V_solution  =  $\frac{{\mathrm{m}}_{\mathrm{NaCl}}}{{\mathrm{M}}_{\mathrm{NaCl}}}$  

⇒ m_NaCl  = M x V_solution x M_NaCl   with M_NaCl = molar mass of NaCl = M_Na + M_Cl = 23.0 + 35.4 = 58.4 g.mol^-1

⇒ m_NaCl  = 0.4 x 0.250 x 58.4 = 5.84 g

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

 In dilution, the amount of solute does not change: the number of moles are the same before and after dilution.

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

In dilution, the amount of solute does not change: n₁ = n₂ ⇒ M₁ x V₁ = M₂ x V₂
2.0 x V₁ = 0.5 x 0.300 ⇒ V₁ = 0.075 L = 75 mL

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