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2 years ago

The reaction of iron (III) oxide with carbon monoxide produces iron and carbon dioxide.

Chemistry

1 answer:

sleet_krkn [62]2 years ago

7 0

Answer:

6.65 grams of Fe₂O₃ are required to produce 4.65g Fe.

Explanation:

The balanced reaction is:

Fe₂O₃ + 3 CO → 2 Fe + 3 CO₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Fe₂O₃: 1 mole
CO: 3 moles
Fe: 2 moles
CO₂: 3 moles

The molar mass of each compound is:

Fe₂O₃: 159.7 g/mole
CO: 28 g/mol
Fe: 55.85 g/mole
CO₂: 44 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Fe₂O₃: 1 mole* 159.7 g/mole= 159.7 grams
CO: 3 moles* 28 g/mol= 84 grams
Fe: 2 moles* 55.85 g/mole= 111.7 grams
CO₂: 3 moles* 44 g/mole= 132 grams

Then you can apply the following rule of three: if by stoichiometry 111.7 grams of Fe are produced from 159.7 grams of Fe₂O₃, 4.65 grams of Fe are produced from how much mass of Fe₂O₃?

$mass of Fe_{2} O_{3} =\frac{4.65 grams of Fe*159.7 grams of Fe_{2} O_{3}}{111.7grams of Fe}$

mass of Fe₂O₃= 6.65 grams

<u><em>6.65 grams of Fe₂O₃ are required to produce 4.65g Fe.</em></u>

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Answer:

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<h3><u>Saturated fat is found in:</u></h3>

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3 years ago

A 19.0 g piece of metal is heated to 99.0 °C and then placed in 150 mL of water at 21°C. The temperature of the water rose to 23

Anna35 [415]

Answer:

$0.8696\ \text{J/g}^{\circ}\text{C}$

Explanation:

$m_m$ = Mass of metal = 19 g

$c_m$ = Specific heat of the metal

$\Delta T_m$ = Temperature difference of the metal = $99-23=76^{\circ}\text{C}$

V = Volume of water = 150 mL = $150\ \text{cm}^3$

$\rho$ = Density of water = $1\ \text{g/cm}^3$

$c_w$ = Specific heat of the water = 4.186 J/g°C

$\Delta T_w$ = Temperature difference of the water = $23-21=2^{\circ}\text{C}$

Mass of water

$m_w= \rho V\\\Rightarrow m_w=1\times 150\\\Rightarrow m_w=150\ \text{g}$

Heat lost will be equal to the heat gained so we get

$m_mc_m\Delta T_m=m_wc_w\Delta T_w\\\Rightarrow c_m=\dfrac{m_wc_w\Delta T_w}{m_m\Delta T_m}\\\Rightarrow c_m=\dfrac{150\times 4.186\times 2}{19\times 76}\\\Rightarrow c_m=0.8696\ \text{J/g}^{\circ}\text{C}$

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2 years ago

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2 years ago

(e) Another student investigated the rate of a different reaction

vodka [1.7K]

Answer:

0.07 g/s.

Explanation:

From the question given above, the following data were obtained:

Mass lost = 9.85 g

Time taken = 2 min 30 s

Mean rate =?

Next, we shall convert 2 min 30 s to seconds (s). This can be obtained as follow:

1 min = 60 s

Thus,

2 min = 2 × 60 = 120 s

Therefore,

2 min 30 s = 120 s + 30 s = 150 s

Finally, we shall determine the mean rate of the reaction. This can be obtained as illustrated below:

Mass lost = 9.85 g

Time taken = 150 s

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Mean rate = mass lost / time taken

Mean rate = 9.85 / 150

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