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1 year ago

The thermite reaction reacts iron (III) oxide, Fe2O3, with aluminium powder,Al, the form aluminium oxide, Al2O3 and iron, Fe.

Chemistry

1 answer:

lara [203]1 year ago

5 0

Answer:

This answer assumes that the strated "16.0g of iron" was meant to be 16.0 grams of iron(III) oxide.

Explanation:

To start, the thermite equation must be balanced.

I find:

1Fe2O3 + 2Al = 1Al2O3 + 2Fe

This tells us we need 2 moles of Al for every 1 mole of Fe2O3.

Now calculate the moles of each reactant:

Moles Fe2O3: 16.0 g/159.7 g/mole = 0.100 moles Fe2O3

Moles Al: 8.1 /26.98 g/mole = 0.300 moles Al

The balanced equation says that in order to react all of the Fe2O3 we'd need twice that amount (in moles) of the Al. (0.100 moles Fe2O3)*(2) = 0.200 moles Al.

Which of the two reactants is the limiting reagent?

We have more than enough moles of Al to react with 0.10 moles of Fe2O3. (We have 0.300 moles Al and all we need is 0.200 moles to react with the 0.10 moles of Fe2O3. Fe2O3 is the limiting reagent.

Calculate the maximum mass of iron of iron that could be formed using these quantities of reactants.

The balanced equation tells us that we will obtain 2 moles of Fe for every 1 mole of Fe2O3 consumed. Since Fe2O3 is the limiting reagent, we will assume that it completely reacts. That means 0.1 moles of Fe2O3 is reacted. Since we expect twice that many moles of Fe, we should obtain 0.200 moles of Fe. At 55.85 g/mole, we should obtain:

(0.200 moles Fe)*(55.85 g Fe/mole Fe) = 11.2 grams Fe

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Answer:

A. 2C + H₂ ⟶ CH₄

Explanation:

A. 2C + H₂ ⟶ CH₄

UNBALANCED. 2C on the left and 1C on the right

B. 2Al₂O₃ ⟶ 4Al + 3O₂

Balanced. Same number of each type of atom on each side.

C. 2H₂O₂ ⟶ 2H₂O + O₂

Balanced. Same number of each type of atom on each side.

D. 2C₂H₆ + 7O₂ ⟶ 4CO₂ + 6H₂O

Balanced. Same number of each type of atom on each side.

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3 years ago

What happens when soluble fiber is mixed with water?

storchak [24]

Answer:

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3 years ago

If the volume of a gas container at 32 degrees Celsius changes from 1.55 L to 755 mL, what will the final temperature be?

QveST [7]

So to solve this you need to know Charles’s law which is: V1/T1=V2/T2. Where T1 and V1 is the initial volume and Temperature and V2 and T2 is the temperature and volume afterwards. So first plug in the numbers you are given. V1= 1.55L T1= 32C° V2= 755mL T2=?. Since your volumes are two different units you change 755mL to be in L so that would be 0.755 L. And since your temp isn’t in Kelvin you do 273+32= 305K°. You then would rearrange your equation to solve for T2 which is V2T1/V1. Then you plug in your numbers (0.755L)(305K)/1.55L. Then you solve and would be 148.5645161 —> 1.49 x 10^2 K

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3 years ago

En un vaso de precipitado de un litro se coloca exactamente 500 mL de agua destilada a temperatura ambiente y se realizan dos ex

Serga [27]

Answer:

1) La masa del agua a temperatura ambiente es de 500 gramos, 2) La masa del agua cuando se congela es de 500 gramos, 3) La masa de agua que queda después de la evaporación es de 400 gramos, 4) Se ha evaporado 100 gramos de agua.

Explanation:

1) ¿Cuál es la masa de agua a temperatura ambiente?

Podemos determinar la masa inicial del agua ( $m_{o}$ ), medido en gramos, al conocer su densidad ( $\rho_{w}$ ), medida en gramos por mililitro, y volumen inicial ocupado en el vaso de precipitado ( $V_{o}$ ), medido en mililitros, a partir de la siguiente expresión:

$m_{o} =\rho_{w}\cdot V_{o}$

Si sabemos que $\rho_{w} = 1\,\frac{g}{mL}$ y $V_{o} = 500\,mL$ , entonces:

$m_{o} = \left(1\,\frac{g}{mL} \right)\cdot (500\,mL)$

$m_{o} = 500\,g$

La masa del agua a temperatura ambiente es de 500 gramos.

2) ¿Cuál es la masa de agua cuando se congela?

Puesto que el proceso de congelación no implica transferencia de masa, la masa de agua se conserva al transformarse en hielo. Por tanto, la masa resultante es de 500 gramos.

3) ¿Cuál es la masa de agua que queda después de la evaporación?

Durante la evaporación una parte del agua es transferida al aire, entonces podemos calcular la masa final ( $m_{f}$ ), medido en gramos, de la sustancia al multiplicar el volumen final ( $V_{f}$ ), medido en mililitros, por la densidad del agua ( $\rho_{w}$ ), medida en gramos por mililitro,. Es decir,

$m_{f} =\rho_{w}\cdot V_{f}$

Si sabemos que $\rho_{w} = 1\,\frac{g}{mL}$ y $V_{f} = 400\,mL$ , entonces:

$m_{f} = \left(1\,\frac{g}{mL} \right)\cdot (400\,mL)$

$m_{f} = 400\,g$

La masa de agua que queda después de la evaporación es de 400 gramos.

4) ¿Qué masa de agua se evaporó?

Determinamos que la masa evaporada de agua ( $m_{v}$ ), medida en gramos, es igual a la diferencia entre las masas inicial y final, ambas medidas en gramos:

$m_{v} =m_{o}-m_{f}$

Si $m_{o} = 500\,g$ y $m_{f} = 400\,g$ , entonces tenemos que:

$m_{v} = 500\,g -400\,g$

$m_{v} = 100\,g$

Se ha evaporado 100 gramos de agua.

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3 years ago

You need 455 ml of a 75% alcohol solution. on hand, you have a 25% alcohol mixture. you also have a 90% alcohol mixture. how muc

marishachu [46]

X ml - 25% alcohol mixture
y ml - 90% alcohol mixture

x+y = 455

0.25x ml alcohol in x ml of 25% alcohol mixture
0.9y ml alcohol in y ml of 90% alcohol mixture
0.75*455= 341.25 ml alcohol in 455 ml of 75% alcohol mixture

0.25x+0.9y=341.25

System of equations:

x+y = 455 /*(-0.25) ------> -0.25x-0.25y = -0.25*455
0.25x+0.9y=341.25

-0.25x-0.25y=-113.75
0.25x+0.9y=341.25 Add both equations

0.25x+0.9y-0.25x-0.25y=341.25-113.75
0.65y =227.5
y=227.5/0.65 = 350 ml of 90% alcohol mixture

x+y=455
x+355=455
x=100 ml of 25% alcohol mixture

8 0

3 years ago