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3 years ago

Fe2O3+2Al=Al2O3+2Fe

2 answers:

6 0

We determine the limiting reactant by using the moles present in the equation and the actual moles.
According to equation, ratio of Fe₂O₃ : Al = 1 : 2
Actual moles of Fe₂O₃ = 187.3 / (56 x 2 + 16 x 3)
= 1.17
Actual moles of Al = 94.51 / 27
= 3.5
Fe₂O₃ is limiting. Fe₂O₃ required:
(moles Al)/2 = 3.5/2 = 1.75
Moles to be added = 1.75 - 1.17
= 0.58
Mass to be added = moles x Mr
= 0.58 x (56 x 2 + 16 x 3)
= 92.8 grams

Studentka2010 [4]3 years ago

4 0

92.41g, this is the correct answer

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Answer:

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3 years ago

A chemistry student must write down in her lab notebook the concentration of a solution of sodium hydroxide. The concentration o

HACTEHA [7]

Answer:

1.099 gmL¯¹ ≈ 1.1 gmL¯¹

Explanation:

From the question given above, the following were obtained:

Mass of empty cylinder = 9.5 g

Mass Cylinder + NaOH = 31.92 g

Volume of solution = 20.4 mL

Concentration of solution =?

Next, we shall determine the mass of sodium hydroxide, NaOH. This can be obtained as as illustrated below:

Mass of empty cylinder = 9.5 g

Mass Cylinder + NaOH = 31.92 g

Mass of NaOH =?

Mass of NaOH = (Mass Cylinder + NaOH) – (Mass of empty cylinder)

Mass of NaOH = 31.92 – 9.5

Mass of NaOH = 22.42 g

Finally, we shall determine concentration of the solution as follow:

Mass of NaOH = 22.42 g

Volume of solution = 20.4 mL

Concentration of solution =?

Concentration = mass /volume

Concentration of solution = 22.42 / 20.4

Concentration of solution = 1.099 gmL¯¹ ≈ 1.1 gmL¯¹

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4 years ago

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3 years ago

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Answer:

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2 years ago

1‑Propanol ( P⁰ 1 = 20.9 Torr at 25 ⁰C ) and 2‑propanol ( P⁰ 2 = 45.2 Torr at 25 ⁰C ) form ideal solutions in all proportions. L

anastassius [24]

Answer : The mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

Explanation : Given,

Vapor presume of 1‑Propanol $(P^o_1)$ = 20.9 torr

Vapor presume of 2‑Propanol $(P^o_2)$ = 45.2 torr

Mole fraction of 1‑Propanol $(x_1)$ = 0.540

Mole fraction of 2‑Propanol $(x_2)$ = 1-0.540 = 0.46

First we have to calculate the partial pressure of 1‑Propanol and 2‑Propanol.

$p_1=x_1\times p^o_1$

where,

$p_1$ = partial vapor pressure of 1‑Propanol

$p^o_1$ = vapor pressure of pure substance 1‑Propanol

$x_1$ = mole fraction of 1‑Propanol

$p_1=(0.540)\times (20.9torr)=11.3torr$

and,

$p_2=x_2\times p^o_2$

where,

$p_2$ = partial vapor pressure of 2‑Propanol

$p^o_2$ = vapor pressure of pure substance 2‑Propanol

$x_2$ = mole fraction of 2‑Propanol

$p_2=(0.46)\times (45.2torr)=20.8torr$

Thus, total pressure = 11.3 + 20.8 = 32.1 torr

Now we have to calculate the mole fraction of vapor phase 1‑Propanol and 2‑Propanol.

$\text{Mole fraction of 1-Propanol}=\frac{\text{Partial pressure of 1-Propanol}}{\text{Total pressure}}=\frac{11.3}{32.1}=0.352$

and,

$\text{Mole fraction of 2-Propanol}=\frac{\text{Partial pressure of 2-Propanol}}{\text{Total pressure}}=\frac{20.8}{32.1}=0.648$

Thus, the mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

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3 years ago