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3 years ago

How many moles of Al are necessary to form 45.0 g of AlBr, from this

Chemistry

1 answer:

drek231 [11]3 years ago

3 0

Answer:

0.169 mole of Al

Explanation:

We'll begin by by calculating the number of mole in 45 gof AlBr₃. This can be obtained as follow:

Mass of AlBr₃ = 45 g

Molar mass of AlBr₃ = 27 + (3×80)

= 27 + 240

= 267 g/mol

Mole of AlBr₃ =?

Mole = mass /molar mass

Mole of AlBr₃ = 45 / 267

Mole of AlBr₃ = 0.169 mole

Finally, we shall determine the number of mole of Al needed to produce 45 g (i.e 0.169 mole) of AlBr₃. This can be obtained as illustrated below:

2Al + 3Br₂ —> 2AlBr₃

From the balanced equation above,

2 moles of Al reacted to produce 2 moles of AlBr₃.

Therefore, 0.169 mole of Al will also react to produce 0.169 mole of AlBr₃.

Thus, 0.169 mole of Al is needed for the reaction.

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Find the empirical formula of the following compounds:

Aneli [31]

The empirical formula of the following compounds 0.903 g of phosphorus combined with 6.99 g of bromine.

<h3>What is empirical formula?</h3>

The simplest whole number ratio of atoms in a compound is the empirical formula of a chemical compound in chemistry. Sulfur monoxide's empirical formula, SO, and disulfur dioxide's empirical formula, S2O2, are two straightforward examples of this idea. As a result, both the sulfur and oxygen compounds sulfur monoxide and disulfur dioxide have the same empirical formula.

<h3>How to find the empirical formula?</h3>

Convert the given masses of phosphorus and bromine into moles by multiplying the reciprocal of their molar masses. The molar masses of phosphorus and bromine are 30.97 and 79.90 g/mol, respectively.

Moles phosphorus = 0.903 g phosphorus $\frac{mol phosphorus}{ 30.97 g phosphorus}$ = 0.0293 mol

Moles bromine 6.99 g bromine $\frac{mol bromine}{79.90 g bromine}$ =0.0875 mol

The preliminary formula for compound is P0.0293Bro.0875. Divide all the subscripts by the subscript with the smallest value which is 0.0293. The empirical formula is P1.00Br2.99 ≈ P₁Br3 or PBr3

To learn more about empirical formula visit:

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8 0

1 year ago

Find the mass of an object that has a density of 1.5 g/cm3 and has a volume of 8cm3

zubka84 [21]

Answer:

density= 1.5 g/cm3

volume= 8cm 3

mass = density×volume

mass= 1.5×8

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7 0

3 years ago

If volumes are additive and 253 mL of 0.19 M potassium bromide is mixed with 441 mL of a potassium dichromate solution to give a

Alexxx [7]

Answer:

The concentration of the Potassium Dichromate solution is 0.611 M

Explanation:

First of all, we need to understand that in the final solution we'll have potassium ions coming from KBr and also K2Cr2O7, so we state the dissociation equations of both compounds:

KBr (aq) → K+ (aq) + Br- (aq)

K2Cr2O7 (aq) → 2K+ (aq) + Cr2O7 2- (aq)

According to these balanced equations when 1 mole of KBr dissociates, it generates 1 mole of potassium ions. Following the same thought, when 1 mole of K2Cr2O7 dissociates, we obtain 2 moles of potassium ions instead.

Having said that, we calculate the moles of potassium ions coming from the KBr solution:

0.19 M KBr: this means that we have 0.19 moles of KBr in 1000 mL solution. So:

1000 mL solution ----- 0.19 moles of KBr

253 mL solution ----- x = 0.04807 moles of KBr

As we said before, 1 mole of KBr will contribute with 1 mole of K+, so at the moment we have 0.04807 moles of K+.

Now, we are told that the final concentration of K+ is 0.846 M. This means we have 0.846 moles of K+ in 1000 mL solution. Considering that volumes are additive, we calculate the amount of K+ moles we have in the final volume solution (441 mL + 253 mL = 694 mL):

1000 mL solution ----- 0.846 moles K+

694 mL solution ----- x = 0.587124 moles K+

This is the final quantity of potassium ion moles we have present once we mixed the KBr and K2Cr2O7 solutions. Because we already know the amount of K+ moles that were added with the KBr solution (0.04807 moles), we can calculate the contribution corresponding to K2Cr2O7:

0.587124 final K+ moles - 0.04807 K+ moles from KBr = 0.539054 K+ moles from K2Cr2O7

If we go back and take a look a the chemical reactions, we can see that 1 mole of K2Cr2O7 dissociates into 2 moles of K+ ions, so:

2 K+ moles ----- 1 K2Cr2O7 mole

0.539054 K+ moles ---- x = 0.269527 K2Cr2O7 moles

Now this quantity of potassium dichromate moles came from the respective solution, that is 441 mL, so we calculate the amount of them that would be present in 1000 mL to determine de molar concentration:

441 mL ----- 0.269527 K2Cr2O7 moles

1000 mL ----- x = 0.6112 K2Cr2O7 moles = 0.6112 M

6 0

3 years ago

How many neutrons are in selenium -36?

Soloha48 [4]

Selenium (Se) the most common isotope of this element. The nucleus consists of 34 protons (red) and 46 neutrons (blue).

5 0

3 years ago

A tablet of Pain Be Gone Aspirin, which had a mass of 1.213 g, was pulverized and 1.159 g were dissolved in 10.0 mL of ethyl alc

Elan Coil [88]

Answer:

a. Moles of $NaOH$ = 0.001643 moles

b. 0.296 g

c. 0.3098 g

d. Not acceptable

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For $NaOH$ :

Molarity = 0.1052 M

Volume = 15.62 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 15.62×10⁻³ L

Thus, moles of $NaOH$ :

$Moles=0.1052 \times {15.62\times 10^{-3}}\ moles$

Moles of $NaOH$ = 0.001643 moles

The reaction of NaOH with the acetylsalicylic acid is in the ratio of 1:1.

Thus, Moles of NaOH = Moles of acetylsalicylic acid = 0.001643 moles

Molar mass of acetylsalicylic acid = 180.16 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

Mass = Moles * Molar mass = 0.001643 moles * 180.16 g/mol = 0.296 g

1.159 g of sample contains 0.296 g of acetylsalicylic acid

1.213 g of sample contains $\frac{0.296}{1.159}\times 1.213$ g of acetylsalicylic acid

Mass of acetylsalicylic acid = 0.3098 g = 309.8 mg

d. Sample contains = 309.8 mg

Manufacturer claiming = 315 mg to 335 mg

Thus , it is not acceptable.

5 0

3 years ago