smc-3682-50-Stoichiometric Calculations

CHEMISTRY, M8 EQ-Bank 23

$\ce{Fe^2^+}$ and $\ce{X} $ react to form an ionic compound according to the general equation

$\ce{aFe^2^+ + $b$(X)\rightleftharpoons [Fe_a(X)_b]^2^a^+}$

where $\ce{$a$}$ and $\ce{$b$}$ are numbers representing the ratio in which $\ce{Fe^2^+}$ and $\ce{X} $ combine.

Spectrophotometry was used to determine the stoichiometric ratio between $\ce{Fe^2^+}$ and $\ce{X} $. To do this, eight 10 mL samples were prepared by reacting solutions of $\ce{Fe^2^+}$ with solutions of $\ce{X} $ in varying ratios. All $\ce{Fe^2^+}$ and $\ce{X} $ solutions had the same concentration. The absorbance of the samples is tabulated below.

On the grid, construct a graph of absorbance against volume of $\ce{Fe^2^+}$ solution from 0.00 mL to 6.00 mL, and draw TWO lines of best fit. (3 marks)

The reaction proceeds according to the general equation
$\ce{aFe^2^+ + $b$X \rightleftharpoons [Fe_a(X)_b]^2^a^+}$.
Find the values of $\ce{$a$}$ and $\ce{$b$}$ . Justify your answer with reference to the data given and the graph in part (a). (3 marks)

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b. Find values of $\ce{$a$}$ and $\ce{$b$}$:

The graph shows that as the volume of compound $\ce{X} $ decreases between 10mL and 7.5mL, there is an increase in $\ce{Fe^{2+}}$ ions. This indicates an excess of $\ce{Fe^{2+}}$ ions limiting the products.
The absorbance reaches a maximum when the 2.5 mL of compound $\ce{Fe^{2+}}$ is added to 7.5 mL of compound $\ce{X}$. Since the concentrations of the initial solutions are equal (given), equal volumes contain equal moles.
The ratio of volumes at peak absorbance = 2.5 : 7.5 = 1 : 3 (i.e. the correct stoichiometric ratio).
Hence $\ce{$a$ = 1}$ and $\ce{$b$ = 3}$.
The curve then turns down sharply, indicating a reduction in $\ce{Fe^{2+}}$.

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b. Find values of $\ce{$a$}$ and $\ce{$b$}$:

The graph shows that as the volume of compound $\ce{X} $ decreases between 10mL and 7.5mL, there is an increase in $\ce{Fe^{2+}}$ ions. This indicates an excess of $\ce{Fe^{2+}}$ ions limiting the products.
The absorbance reaches a maximum when the 2.5 mL of compound $\ce{Fe^{2+}}$ is added to 7.5 mL of compound $\ce{X}$. Since the concentrations of the initial solutions are equal (given), equal volumes contain equal moles.
The ratio of volumes at peak absorbance = 2.5 : 7.5 = 1 : 3 (i.e. the correct stoichiometric ratio).
Hence $\ce{$a$ = 1}$ and $\ce{$b$ = 3}$.
The curve then turns down sharply, indicating a reduction in $\ce{Fe^{2+}}$.

CHEMISTRY, M8 EQ-Bank 12

A colorimeter was used to calculate the percentage of iron in a 0.200 gram tablet. The tablet was dissolved and oxidised, then reacted with thiosulfate according to the equation

$\ce{Fe^3^+($aq$) + SCN^-($aq$)\rightarrow [FeSCN]^2^+($aq$)}$

The resulting solution was made up to 200 mL with distilled water. The absorbance of the final solution was measured to be 0.6105.

The calibration curve shows the absorbance of various concentrations of $\ce{Fe^3^+}$.

Calculate the percentage of iron in the tablet. (3 marks)

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0.307%

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$\ce{Fe^{3+}($aq$) + SCN^-($aq$)\rightarrow [FeSCN]^2^+($aq$)}$

$\ce{\text{Using the graph:}}$

$\ce{Absorbance = 0.615 }$

$\ce{\rightarrow Initial [Fe^{3+}] in 200 mL = 55 \times 10^{-6} = 5.5 \times 10^{-5} mol L^{-1}}$

$\ce{n(Fe^{3+}) = c \times V = 5.5 \times 10^{-5} \times 0.2 = 1.1 \times 10^{-5} mol} $

$\ce{MM(Fe) = 55.85 g mol^{-1}} $

$\ce{m(Fe) = 1.1 \times 10^{-5} \times 55.85 = 6.14 \times 10^{-4} g} $

\[ \ce{\therefore \text{% Fe} = \frac{6.14 \times 10^{-4}}{0.200} \times 100 = 0.307\text{%} } \]

CHEMISTRY, M8 2018 HSC 20 MC

The Winkler method is used to determine the amount of dissolved oxygen in a water sample. The procedure involves the following sequence of reactions.

Step 1. $\ce{2Mn^2+(aq) + O2(g) + 4OH-(aq) -> 2MnO(OH)2(s)}$

Step 2. $\ce{MnO(OH)2(s) + 2I-(aq) + 4H+(aq) -> I2(aq) + Mn^2+(aq) + 3H2O(aq)}$

Step 3. $\ce{I2(aq) + 2S2O3^2-(aq) -> 2I-(aq) + S4O6^2-(aq)}$

When a 5.00 L sample of water was analysed using the Winkler method, a total of $ 4.00 \times 10^{-3}$ mol of thiosulfate $\ce{S2O3^2-}$ was required in Step 3.

What concentration of oxygen was present in the original sample?

`3.20 \ text{mg L}^{-1}`
`6.40 \ text{mg L}^{-1}`
`12.8 \ text{mg L}^{-1}`
`32.0 \ text{mg L}^{-1}`

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`B`

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Consider Step 3:

\[\ce{n(I2) = \frac{1}{2} \times n(S2O3^2-) = \frac{1}{2} \times 4.00 \times 10^{-3} = 2 \times 10^{-3} mol}\]

Consider Step 2:

$\ce{n(MnO(OH)2) = n(I2) = 2 \times 10^{-3} mol}$

Consider Step 1:

\[\ce{n(O2) = \frac{1}{2} \times n(MnO(OH)2) = \frac{1}{2} \times 2.00 \times 10^{-3} = 1 \times 10^{-3} mol}\]

\[\therefore \ce{[O2] = \frac{n}{V} \times MM = \frac{1 \times 10^{-3}}{5.0} \times 32 = 0.0064 g L^{-1}}\]

`=>B`

♦ Mean mark 45%.

CHEMISTRY, M8 2016 HSC 27

The volume of gas formed at 25°C and 100 kPa as hydrochloric acid was added to a pure sample of aluminium is shown in the graph.

Calculate the original mass of the aluminium sample used in the reaction. (4 marks)

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0.109 grams

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\[\ce{Al(s) + 3HCl(aq) -> AlCl3(aq) + \frac{3}{2}H2(g)}\]

The graph shows that all the aluminium has reacted when no more gas is being produced, at a volume of 0.150 L.

\[\ce{n(H2) = \frac{0.150}{24.79} = 0.00605 mol}\]

\[\ce{n(Al) = \frac{2}{3} \times 0.00605 = 0.00403 mol}\]

$\therefore \ce{m(Al) = 0.00403 \times 26.98 = 0.109 g}$

♦ Mean mark 48%.

CHEMISTRY, M8 2019 HSC 20 MC

The manganese content in a 12.0 gram sample of steel was determined by measuring the absorbance of permanganate $\ce{(MnO4^-)} $ using the following process.

The steel sample was dissolved in nitric acid and the $\ce{Mn^2+(aq)}$ ions produced were oxidised to $\ce{MnO4^-(aq)}$ by periodate ions, $\ce{IO4^-(aq)}$ , according to the following equation.

$\ce{2Mn^2+(aq) + 5IO4^-(aq) + 3H2O(l) → 2MnO4^-(aq) + 5IO3^-(aq) + 6H+(aq)} $

The resulting solution was made up to a volume of 1.00 L, then 20.0 mL of this solution was diluted to 100.0 mL. The absorbance at 525 nm of the resulting solution was 0.50.

A calibration curve for $\ce{MnO4^-(aq)}$ was constructed and is shown below.

What was the percentage by mass of manganese in the steel sample?

0.019%
0.096%
0.48%
1.0%

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`C`

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From graph, 0.5 absorbance corresponds to 0.25 mg L ¯¹.

$\ce{[MnO4^-]_{dilute}}$	`=(25 xx 10^(-3))/(54.94+16 xx 4)=2.1019 xx 10^(-4)\ text{mol L}^(-1)`
$\ce{[MnO4^-]_{conc}}$	`=2.1019 xx 10^(-4) xx 1/2=1.05095 xx 10^(-3)\ text{mol L}^(-1)`

$\ce{n(MnO4^-) = c \times V = 1.05095 \times 10^{-3} \times 1 = 1.05095 \times 10^{-3} mol}$

$\ce{n(Mn^2) = n(MnO4^-) = 1.05095 \times 10^{-3} mol}$

$\ce{m(Mn^2+)}=\text{n} \times \text{MM}=1.05095 \times 10^{-3} \times 54.94=5.774 \times 10^{-3} \text{g} $

`:.\ text{Mn}^(2+)text{(%)} = (5.774 xx 10^(-3))/12.0 xx 100text(%) = 0.48text(%)`

`=>C`

♦♦ Mean mark 43%.