# Quiz 1 - Chemical Calculations | Chemical Calculations

1

What is the mass of 1 mole of benzaldehyde (C6H5CHO)?

nbenzaldehyde  =  $\frac{{\mathrm{m}}_{\mathrm{benzaldehyde}}}{{\mathrm{M}}_{\mathrm{benzaldehyde}}}$   ⇒   mbenzaldehyde = nbenzaldehyde x Mbenzaldehyde

Mbenzaldehyde = 7 x MC + 6 x MH + MO = 7 x 12.0 + 6 x 1.0 + 16.0 = 106.0 g.mol-1

⇒ mbenzaldehyde = 1 x 106.0 = 106 g

2

What is the mass of carbon present in 110 g of carbon dioxide?

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

MC = 12.0 g.mol-1;   MCO2 = MC + 2 x MO = 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 =  x mass of CO2 = 0.273 x 110 = 30.0 g

3

What is the molarity of a solution that contains 8 moles of solute in 2 liters of solution?

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

4

How many moles does a balloon filled with 10 g of hydrogen have?

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

There are 2 moles of H in 1 mole of H2   ⇒   nH2 = ${\mathrm{n}}_{\mathrm{H}}}{2}$ = 4.95 moles

5

What is the mass of NaCl in 250 mL of a 0.4 M solution?

number of moles nNaCl  =  M x Vsolution  =  $\frac{{\mathrm{m}}_{\mathrm{NaCl}}}{{\mathrm{M}}_{\mathrm{NaCl}}}$

⇒ mNaCl  = M x Vsolution x MNaCl   with MNaCl = molar mass of NaCl = MNa + MCl = 23.0 + 35.4 = 58.4 g.mol-1

⇒ mNaCl  = 0.4 x 0.250 x 58.4 = 5.84 g

6

How many of carbon atoms do you have in 0.4 mole of C2H4?

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

There are 2 moles of C in 1 mole of C2H4: nC = 2 x nC2H4

⇒ N = nC x NA = 2 x nC2H4 x NA = 2 x 0.4 x 6.022 x 1023 = 4.82 x 1023 atoms

7

What does not change when a solution is diluted by the addition of solvent?

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

8

What is the limiting reagent in a chemical reaction?

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

9

How many mL of a 2.0 M NaBr solution are needed to make 300.0 mL of 0.5 M NaBr?

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

10

How many grams of NH3 can be prepared from 5 g of N2 and 5 g of H2?

nN2$\frac{{\mathrm{m}}_{\mathrm{N}2}}{{\mathrm{M}}_{\mathrm{N}2}}$$\frac{5.0}{28.0}$ = 0.179 mole;   nH2 = $\frac{{\mathrm{m}}_{\mathrm{H}2}}{{\mathrm{M}}_{\mathrm{H}2}}$ = $\frac{5.0}{2.02}$ = 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

nN2 < $\frac{{\mathrm{n}}_{\mathrm{H}2}}{3}$ ⇒ N2 is the limiting reagent and nNH3 = 2 x nN2 = 0.357 mole

mNH3 = nNH3 x MNH3 = 0.357 x 17.0 = 6.08 g