All categories

3 years ago

How many moles of aluminum are needed to react completely with 1.2 mol of feo

Chemistry

1 answer:

Lelu [443]3 years ago

3 0

Answer:

0.80 mol Al

Explanation:

1) Word reaction:

Aluminum + ferrous oxide → aluminum oxide + iron

This is a single replacement reaction, in which aluminum, a more acitve metal than iron, replaces the iron in the ferrous oxide to form aluminum oxide and iron.

2) Skeleton chemical equation:

Al + FeO → Al₂O₃ + Fe

3) Balanced chemical equation:

Add the coeffcients to comply with the law of conservation of mass:

2Al + 3FeO → Al₂O₃ + 3Fe

4) Mole ratio of Al and FeO:

2 mol Al : 3 mol FeO

5) Set a proportion with the unkown and solve:

x / 1.2 mol FeO = 2 mol Al / 3 mol FeO

x = 1.2 mol FeO × 2 mol Al / 3 mol FeO = 0.80 mol Al ← answer

You might be interested in

The flask contains 10.0 mL of HCl and a few drops of phenolphthalein indicator. The buret contains 0.160 M NaOH. It requires 18.

olchik [2.2K]

Answer:

Approximately $0.291\; \rm M$ (rounded to two significant figures.)

Explanation:

The unit of concentration $\rm M$ is the same as $\rm mol \cdot L^{-1}$ (moles per liter.) On the other hand, the volume of both the $\rm NaOH$ solution and the original $\rm HCl$ solution here are in milliliters. Convert these two volumes to liters:

$V(\mathrm{NaOH}) = 18.2\; \rm mL = 18.2 \times 10^{-3}\; \rm L = 0.0182\; \rm L$ .
$V(\text{$\mathrm{HCl}$, original}) = 10.0\; \rm mL = 10.0\times 10^{-3}\; \rm L = 0.0100\; \rm L$ .

Calculate the number of moles of $\rm NaOH$ in that $0.0182\; \rm L$ of $0.160\; \rm M$ solution:

$\begin{aligned} n(\mathrm{NaOH}) &= c(\mathrm{NaOH})\cdot V(\mathrm{NaOH})\\ &= 0.160\; \rm mol \cdot L^{-1} \times 0.0182\; \rm L \approx 0.00291\; \rm mol\end{aligned}$ .

$\rm HCl$ reacts with $\rm NaOH$ at a one-to-one ratio:

$\rm HCl\; (aq) + NaOH\; (aq) \to NaCl\; (aq) + H_2O\; (l)$ .

Coefficient ratio:

$\displaystyle \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} = 1$ .

In other words, one mole of $\rm NaOH$ would neutralize exactly one mole of $\rm HCl$ . In this titration, $0.291\; \rm mol$ of $\rm NaOH\!$ was required. Therefore, the same amount of $\rm HC$ should be present in the original solution:

$\begin{aligned}&n(\text{$\mathrm{HCl}$, original})\\ &= n(\mathrm{NaOH})\cdot \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} \\ &\approx 0.00291\; \rm mol \times 1 = 0.00291\; \rm mol\end{aligned}$ .

Calculate the concentration of the original $\rm HCl$ solution:

$\displaystyle c(\text{$\mathrm{HCl}$, original}) = \frac{n(\text{$\mathrm{HCl}$, original})}{V(\text{$\mathrm{HCl}$, original})} \approx \frac{0.00291\; \rm mol}{0.0100\; \rm L} \approx 0.291\; \rm M$ .

5 0

3 years ago

In what element will metallic bonding occur?

Hatshy [7]

D

A- Harvard university professor

6 0

3 years ago

Help! Drag and drop the answer into the box to match each method of how to separate solutions to its description.

Flura [38]

Answer:

1) D 2) A 3) C

Explanation:

7 0

3 years ago

What is the chemical formula for salt?

Hoochie [10]

In terms of the most common type of salt, sodium chloride, NaCl is the chemical formula of this salt,

3 0

3 years ago

If total pressure of multiple gases is 512 mmHg, and the pressure of oxygen gas is 332 mmHg and the pressure of carbon monoxide

deff fn [24]

Answer: 67 mmHg

Explanation:

According to Dalton's Gas Law, the total pressure of a mixture of gases is the sum of the pressure of each individual gas.

i.e Ptotal = P1 + P2 + P3 + .......

In this case,

Ptotal = 512 mmHg

P(oxygen) = 332 mmHg

P(carbon mono-oxide) = 113 mmHg

Remaining pressure (P3) = ?

To get P3, apply Dalton's Gas Law formula

Ptotal = P(oxygen) + P(carbon mono-oxide) + P3

512 mmHg = 332 mmHg + 113 mmHg + P3

512 mmHg = 445 mmHg + P3

P3 = 512 mmHg - 445 mmHg

P3 = 67 mmHg

Thus, the remaining pressure is 67 mmHg

5 0

3 years ago