Why is electron arrangement so important

If you had excess aluminum, how many moles of aluminum chloride could be produced from 21.0 g of chlorine gas, cl2?

AleksandrR [38]

When aluminum metal is made to contact with chlorine gas (Cl₂), a highly exothermic reaction proceeds. This produces aluminum chloride (AlCl₃) powder. The balanced chemical equation for this reaction is shown below:

2Al(s) + 3Cl₂(g) → 2AlCl₃(s)

Since it was stated that aluminum is in excess, this means that the amount of AlCl₃ produced will only depend on the amount of Cl₂ gas available. The molar mass of Cl₂ is 70.906 g/mol. Using stoichiometry, we have the following equation:

(21.0 g Cl₂/ 70.906 g/mol Cl₂) x 2 mol AlCl₃/ 2 mol Cl₂ = 0.1974 mol AlCl₃

Thus, we have determined that 0.1974 <span>moles of aluminum chloride can be produced from 21.0 g of chlorine gas. </span>

3 0

3 years ago

Match each chemical reaction with the type of reaction that best describes it.

r-ruslan [8.4K]

2 Al + 6 HCl → 2 AlCl₃ + 3 H₂ (single displacement)

Ca + Br₂ → CaBr₂ (synthesis)

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O (combustion)

2 NaCl → 2 Na + Cl₂ (decomposition)

FeS + 2 HCl → FeCl₂ + H₂S (double displacement)

single displacement - is a chemical reaction of the following type: A + BC → AC + B

double displacement - is a chemical reaction of the following type: AB + CD → AC + BD

synthesis - the chemical product is obtained by combining in a synthesis the constituent elements

combustion - usually a exothermic reaction of a particular compound with oxygen

decomposition - degradation of a compound in simpler elements

8 0

3 years ago

Calculate the average rate of decomposition of NH4NO2 by the reaction 2NH4O5(aq) --> N2(g) H2O(g) at the following time inter

igor_vitrenko [27]

Answer:

$r=-0.033\frac{M}{s}$

Explanation:

Hello!

In this case, since the average rate of reaction is computed as a change given by:

$r=\frac{\Delta [NH_4NO_2 ]}{\Delta t}$

In such a way, given the concentrations at the specified times, we plug them in to obtain:

$r=\frac{(1.48M-4.12M)}{(330s-250s)}\\\\r=-0.033\frac{M}{s}$

Whose negative sign means the concentration decreased due to the decomposition.

Best regards!

6 0

3 years ago

What's a household item that has electrical energy

olga nikolaevna [1]

Answer:

Washing machine.

Dryer.

Television.

Cell phone.

Explanation:

8 0

2 years ago

Read 2 more answers

In science, we like to develop explanations that we can use to predict the outcome of events and phenomena. Try to develop an ex

Kay [80]

The question is incomplete. The complete question is :

In science, we like to develop explanations that we can use to predict the outcome of events and phenomena. Try to develop an explanation that tells how much NaOH needs to be added to a beaker of HCl to cause the color to change. Your explanation can be something like: The color change will occur when [some amount] of NaOH is added because the color change occurs when [some condition]. The goal for your explanation is that it describes the outcome of this example, but can also be used to predict the outcome of other examples of this phenomenon. Here's an example explanation: The color of the solution will change when 40 ml of NaOH is added to a beaker of HCl because the color always changes when 40ml of base is added. Although this explanation works for this example, it probably won't work in examples where the flask contains a different amount of HCl, such as 30ml. Try to make an explanation that accurately predicts the outcome of other versions of this phenomenon.

Solution :

Consider the equation of the reaction between NaOH and $HCl$

NaOH (aq) + HCl (aq) → NaCl(aq) + $H_2O (l)$

The above equation tells us that $1 \text{mole}$ of $NaOH$ reacts with $1 \text{mole}$ of $HCl$ .

So at the equivalence point, the moles of NaOH added = moles of $HCl$ present.

If the volume of the $HCl$ taken = $V_1$ mL and the conc. of $HCl$ = $M_1$ mole/L

The volume of NaOH added up to the color change = $V_2 \text{ and conc of NaOH = M}_2$ mole/L

Moles of $HCl$ taken = $V_1 \ mL \times M_1 \ mol/100 \ mL = V_2M_2 \times 10^{-3}$ moles.

The color change will occur when the moles of NaOH added is equal to the moles of $HCl$ taken.

Thus when $V_1 M_1 \times 10^{-3} = V_2M_2 \times 10^{-3}$

or when $V_1M_1 = V_2M_2$

or $V_2=\frac{V_1M_1}{M_2}$ mL of NaOH added, we observe the color change.

Where $V_1, M_1$ are the volume and molarity of the $HCl$ taken.

$M_2$ is the molarity of NaOH added.

When both the NaOH and $HCl$ are of the same concentrations, i.e. if $M_1=M_2$ , then $V_2=V_1$

Or the 40 mL of $HCl$ will need 40 mL of NaOH for a color change and

30 mL of $HCl$ would need 30 mL of NaOH for the color change (provided the concentration $M_1=M_2$ )

7 0

3 years ago