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LuckyWell [14K]
3 years ago
8

Question 1(Multiple Choice Worth 4 points)

Chemistry
2 answers:
lys-0071 [83]3 years ago
4 0

Answer:

Question 3: 4/2

Explanation:

I just took the test. ;)

zubka84 [21]3 years ago
3 0

<u>Answer </u>

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.

<u>Solution </u>

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


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