# Quiz | Chemical Calculations

1

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

1 g

106 g

6.022 x 1023 g

Same as 1 mole of benzene (C6H6)

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?

30 g

36.7 g

47.1 g

110 g

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

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

10 moles of H2

9.90 moles of H2

4.95 moles of H2

20.2 moles of H2

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

4

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

1.35 x 1025 atoms

2.41 x 1023 atoms

1.72 x 1022 atoms

4.82 x 1023 atoms

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

5

What is the limiting reagent in a chemical reaction?

The reagent which is completely consumed

The reagent which has the smallest molar mass

The reagent which has the smallest coefficient

All of the above

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

6

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

3.04 g of NH3

6.08 g of NH3

14.05 g of NH3

28.10 g of NH3

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
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