Tarnished silver can be cleaned with toothpaste. Give reason

27.4 g of Aluminum nitrite and 169.9 g of ammonium chloride react to form aluminum chloride, nitrogen, and water. How many grams

GarryVolchara [31]

<u>Answer:</u> The mass of excess reagent (ammonium chloride) remained after the reaction is 62.7 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For aluminium nitrite:</u>

Given mass of aluminium nitrite = 27.4 g

Molar mass of aluminium nitrite = 41 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium nitrite}=\frac{27.4g}{41g/mol}=0.668mol$

<u>For ammonium chloride:</u>

Given mass of ammonium chloride = 169.9 g

Molar mass of ammonium chloride = 53.5 g/mol

Putting values in equation 1, we get:

$\text{Moles of ammonium chloride}=\frac{169.9g}{53.5g/mol}=3.176mol$

The chemical equation for the reaction of aluminium nitrite and ammonium chloride follows:

$Al(NO_2)_3+3NH_4Cl\rightarrow AlCl_3+3N_2+6H_2O$

By Stoichiometry of the reaction:

1 mole of aluminium nitrite reacts with 3 moles of ammonium chloride

So, 0.668 moles of aluminium nitrite will react with = $\frac{3}{1}\times 0.668=2.004mol$ of ammonium chloride.

As, given amount of ammonium chloride is more than the required amount. So, it is considered as an excess reagent.

Thus, aluminium nitrite is considered as a limiting reagent because it limits the formation of product.

Excess moles of ammonium chloride = (3.176 - 2.004) mol = 1.172 moles

Calculating the mass of ammonium chloride by using equation 1, we get:

Excess moles of ammonium chloride = 1.172 moles

Molar mass of ammonium chloride = 53.5 g/mol

Putting values in equation 1, we get:

$1.172mol=\frac{\text{Mass of ammonium chloride}}{53.5g/mol}\\\\\text{Mass of ammonium chloride}=(1.172mol\times 53.5g/mol)=62.7g$

Hence, the mass of excess reagent (ammonium chloride) remained after the reaction is 62.7 grams

3 0

3 years ago

Why does ice have a different density than liquid water?

gulaghasi [49]

Answer:

because it has lighter molecules.

4 0

2 years ago

How could you “supersaturate” solutions, exceeding the amount of dissolved solute possible

Bingel [31]

Answer:

Explanation:

In a saturated solution, more solute cannot be dissolved at a given temperature.

This is because, the solute dissolves in a solvent because of space between particles of solvent but on continuous addition of solute, the space between the solvent particles gets fulfilled. Thus no more solute particle can dissolve in a solvent.

4 0

2 years ago

How many grams of fluorine are contained in 8 molecules of boron trifluoride?

Lelu [443]

<h3>Answer:</h3>

7.57 × 10⁻²² g of F

<h3>Solution:</h3>

Data Given:

Number of Molecules = 8

M.Mass of BF₃ = 67.82 g.mol⁻¹

Mass of Fluorine atoms = ?

Step 1: Calculate Moles of BF₃

Moles = Number of Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

Putting value,

Moles = 8 Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

Moles = 1.33 × 10⁻²³ mol

Step 2: Calculate Mass of BF₃:

Moles = Mass ÷ M.Mass

Solving for Mass,

Mass = Moles × M.Mass

Putting values,

Mass = 1.33 × 10⁻²³ mol × 67.82 g.mol⁻¹

Mass = 9.0 × 10⁻²² g

Step 3: Calculate Mass of Fluorine Atoms:

As,

67.82 g BF₃ contains = 57 g of F

So,

9.0 × 10⁻²² g will contain = X g of F

Solving for X,

X = (9.0 × 10⁻²² g × 57 g) ÷ 67.82 g

X = 7.57 × 10⁻²² g of F

4 0

3 years ago

Read 2 more answers

A chemist titrates 130.0mL of a 0.4248 M lidocaine (C14H21NONH) solution with 0.4429 M HBr solution at 25 degree C . Calculate t

jeka57 [31]

Answer:

pH = 3.36

Explanation:

Lidocaine is a weak base to be titrated with the strong acid HBr, therefore at equivalence point we wil have the protonated lidocaine weak conjugate acid of lidocaine which will drive the pH.

Thus to solve the question we will need to calculate the concentration of this weak acid at equivalence point.

Molarity = mol /V ∴ mol = V x M

mol lidocaine = (130 mL/1000 mL/L) x 0.4248 mol/L = 0.0552 mol

The volume of 0.4429 M HBr required to neutralize this 0.0552 mol is

0.0552 mol x (1L / 0.4429mol) = 0.125 L

Total volume at equivalence is initial volume lidocaine + volume HBr added

0 .130 L +0.125 L = 0.255L

and the concentration of protonated lidocaine at the end of the titration will be

0.0552 mol / 0.255 L = 0.22M

Now to calculate the pH we setup our customary ICE table for weak acids for the equilibria:

protonated lidocaine + H₂O ⇆ lidocaine + H₃O⁺

protonated lidocaine lidocaine H₃O⁺

Initial(M) 0.22 0 0

Change -x +x +x

Equilibrium 0.22 - x x x

We know for this equilibrium

Ka = [Lidocaine] [H₃O⁺] / [protonaded Lidocaine] = x² / ( 0.22 - x )

The Ka can be calculated from the given pKb for lidocaine

Kb = antilog( - 7.94 ) = 1.15 x 10⁻⁸

Ka = Kw / Kb = 10⁻¹⁴ / 1.15 x 10⁻⁸ = 8.71 x 10⁻⁷

Since Ka is very small we can make the approximation 0.22 - x ≈ 0.22

and solve for x. The pH will then be the negative log of this value.

8.71 x 10⁻⁷ = x² / 0.22 ⇒ x = √(/ 8.71 X 10⁻⁷ x 0.22) = 4.38 x 10⁻⁴

( Indeed our approximation checks since 4.38 x 10⁻⁴ is just 0.2 % of 0.22 )

pH = - log ( 4.4x 10⁻⁴) = 3.36

3 0

3 years ago