Question

Question 1(Multiple Choice Worth 4 points)
(05.04 MC)

What mass of CO is needed to react completely with 55.0 g of Fe2O3 in the reaction: Fe2O3(s) + CO(g) → Fe(s) + CO2(g)?

4.82 g CO
9.64 g CO
14.5 g CO
28.9 g CO
Question 2(Multiple Choice Worth 4 points)
(05.04 LC)

Which of the following is a valid mole ratio from the balanced equation 2C3H6 + 9O2 → 6CO2 + 6H2O?

one mole of C three H six over two moles of C O two
six moles of H two O over nine moles of O two
two moles of C three H six over six moles of O two
three moles of H two O over 2 moles of C O two
Question 3(Multiple Choice Worth 4 points)
(05.04 MC)

Read the following chemical equation.

Fe + O2 → Fe2O3

Which of the following fractions can be used for the mole ratio to determine the mass of Fe from a known mass of Fe2O3?

four over two
three over two
two over three
two over four
Question 4(Multiple Choice Worth 4 points)
(05.04 MC)

In an experiment, zinc chlorate decomposed according to the following chemical equation.

Zn(ClO3)2 → ZnCl2 + O2

(Molar mass of Zn(ClO3)2 = 232.29 g/mol; ZnCl2 = 136.286 g/mol; O2 = 31.998 g/mol)

If the mass of zinc chlorate was 150 grams, which of the following calculations can be used to determine the mass of oxygen gas formed?

(150 × 1 × 232.29) ÷ (31.998 × 3) grams
(150 × 3 × 232.29) ÷ (31.998 × 1) grams
(150 × 1 × 31.998) ÷ (232.29 × 3) grams
(150 × 3 × 31.998) ÷ (232.29 × 1) grams
Question 5(Multiple Choice Worth 4 points)
(05.04 LC)

How many moles of sodium cyanide (NaCN) would be needed to produce 4.2 moles of sodium sulfate (Na2SO4)?

H2SO4 + 2NaCN → 2HCN + Na2SO4

2.1 mol NaCN
4.2 mol NaCN
8.4 mol NaCN
12.0 mol NaCN

Answer 1

Answer

Answer 1 : 28.9 g of CO is needed.

Answer 2 : Six moles of $H_{2}O$ over Nine moles of $O_{2}$

Answer 3 : Four over two fraction can be used for the mole ratio to determine the mass of Fe from a known mass of $Fe_{2}O_{3}$.

Answer 4 : Mass of $O_{2}$ = (150 × 3 × 31.998) ÷ (232.29 × 1) grams

Answer 5 : 8.4 moles of sodium cyanide (NaCN) would be needed.

Solution

Solution 1 : Given,

Given mass of $Fe_{2}O_{3}$ = 55 g

Molar mass of $Fe_{2}O_{3}$ = 159.69 g/mole

Molar mass of CO = 28.01 g/mole

Moles of $Fe_{2}O_{3}$ = $\frac{\text{ Given mass of } Fe_{2}O_{3}}{\text{ Molar mass of } Fe_{2}O_{3}}$ = $\frac{55 g}{159.69 g/mole}$ = 0.344 moles

Balanced chemical reaction is,

$Fe_{2}O_{3}(s)+3CO(g)\rightarrow 2Fe(s)+3CO_{2}(g)$

From the given reaction, we conclude that

1 mole of $Fe_{2}O_{3}$ gives              →         3 moles of CO

0.344 moles of $Fe_{2}O_{3}$ gives    →         3 × 0.344 moles of CO

                                                     =         1.032 moles

Mass of CO = Number of moles of CO × Molar mass of CO

                    = 1.032 × 28.01

                    = 28.90 g

Solution 2 : The balanced chemical reaction is,

$2C_{3}H_{6}+9O_{2}\rightarrow 6CO_{2}+6H_{2}O$

From the given reaction, we conclude that the Six moles of $H_{2}O$ over Nine moles of $O_{2}$ is the correct option.

Solution 3 : The balanced chemical reaction is,

$4Fe+3O_{2}\rightarrow 2Fe_{2}O_{3}$

From the given balanced reaction, we conclude that Four over two fraction can be used for the mole ratio to determine the mass of Fe from a known mass of $Fe_{2}O_{3}$.

Solution 4 : Given,

Given mass of $Zn(ClO_{3})_{2}$ = 150 g

Molar mass of $Zn(ClO_{3})_{2}$ = 232.29 g/mole

Molar mass of $O_{2}$ = 31.998 g/mole

Moles of $Zn(ClO_{3})_{2}$ = $\frac{\text{ Given mass of }Zn(ClO_{3})_{2} }{\text{ Molar mass of } Zn(ClO_{3})_{2}}$ = $(\frac{150\times 1}{232.29})moles$

The balanced chemical equation is,

$Zn(ClO_{3})_{2}}\rightarrow ZnCl_{2}+3O_{2}$

From the given balanced equation, we conclude that

1 mole of $Zn(ClO_{3})_{2}$ gives          →       3 moles of $O_{2}$

$(\frac{150\times 1}{232.29})moles$ of $Zn(ClO_{3})_{2}$ gives  →  $[(\frac{150\times 1}{232.29})\times 3] moles$ of $O_{2}$

Mass of $O_{2}$ = Number of moles of $O_{2}$ × Molar mass of  $O_{2}$ = $[(\frac{150\times 1}{232.29})\times 3] \times 31.998 grams$

Therefore, the mass of $O_{2}$ = (150 × 3 × 31.998) ÷ (232.29 × 1) grams

Solution 5 : Given,

Number of moles of $Na_{2}SO_{4}$ = 4.2 moles

Balanced chemical equation is,

$H_{2}SO_{4}+2NaCN\rightarrow 2HCN+Na_{2}SO_{4}$

From the given chemical reaction, we conclude that

1 mole of $Na_{2}SO_{4}$ obtained from 2 moles of NaCN

4.2 moles of $Na_{2}SO_{4}$ obtained   →   2 × 4.2 moles of NaCN

Therefore,

The moles of NaCN needed = 2 × 4.2 = 8.4 moles

Answer 2

The number of moles is obtained by dividing the mass by the molar mass.

1) The reaction equation is;

Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)

Number of moles of Fe2O3  reacted =  55.0 g/160 g/mol =0.34

From the balanced reaction equation;

1 moles of Fe2O3 reacts with 3 moles of CO

0.34 moles of Fe2O3 reacts with  0.34 moles × 3 moles/1 moles

= 1.02 moles of CO

Mass of CO =  1.02 moles × 28 g/mol = 28.9 g CO

2)  Looking at the balanced chemical reaction equation as shown in the question, the valid mole ratio is six moles of H two O over nine moles of O two.

3) The balanced reaction equation is; 4 Fe +3 O2 → 2Fe2O3. From this we can see that the mole ratio to determine the mass of Fe from a known mass of Fe2O3 is four over two.

4) Given that the balanced reaction equation is; Zn(ClO3)2 = ZnCl2 + 3 O2

The equation that could be used to calculate the mass of oxygen formed is;

(150 × 3 × 31.998) ÷ (232.29 × 1) grams

5) The balanced reaction equation is; H2SO4 + 2NaCN = 2HCN + Na2SO4

From the reaction equation;

2 moles of NaCN  yields 1 mole of Na2SO4

4.2 moles of NaCN  yields 4.2 moles ×  1 mole /2 moles

= 2.1 mol NaCN

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