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Determine whether HCl can dissolve each metal sample. If it can, write a balanced chemical reaction showing how the metal dissol

miss Akunina [59]

Answer:

Aluminium can dissolve in HCl

6HCl + 2Al ---> 2AlCl₃ + 3H₂

Minimum volume of HCl required is 76.9 mL.

Explanation:

When a metal dissolves in an acid, a chemical reaction takes place. The chemical reaction occuring is the displacement of the hydrogen ions present in the acid by the metallic ions formed when the metal dissolves in the acid. The displaced hydrogen ion accepts electrons frommthe metal and is evolved as hydrogen gas.

The ability of a metal to be dissolve in an acid with the resultant evolution of hydrogen gas is determined by the position of the metal in the electrochemical series. Metals which are above hydrogen in the electrochemical series are able to displace hydrogen from dilute acids and as such, dissolve in the acid. However, metals which are lower than hydrogen in the electrochemical series are not able to displace hydrogen from dilute acids and as such are not soluble in the acids.

Of the three metals, aluminium, Al, copper, Cu, and silver, Ag, only aluminium is higher than hydrogen in the electrochemical series and as such can dissolve in the 3.25 M HCl.

The equation of the reaction is given below:

6HCl + 2Al ---> 2AlCl₃ + 3H₂

Molar mass of Al = 27 g; Mass of Al reacting = 2.25 g

Number of moles = mass/molar mass

Number of moles of Al present in 2.25 g = 2.25/27 = 0.083 moles

From the equation of reaction, 6 moles of HCL are required to react with 2 moles of Al.

Number of moles of HCl required to react with 0.083 moles of Al = 0.083 × 6/2 = 0.25 moles of HCl

From the formula, molarity = number of moles / volume; volume = number of moles / molarity

Molarity of HCl = 3.25 M; number of molesnof HCl = 0.25 moles

Volume = 0.25 / 3.25 = 0.0769 L or 76.9 mL

Therefore, minimum volume of HCl required is 76.9 mL.

3 0

3 years ago

element m is a metal and its chloride has the formula mcl2. to which group of the periodic table does m most likely belong

Triss [41]

It will belong to the metals because metals bond with nonmetals like chlorine to form ionic compounds

3 0

3 years ago

Read 2 more answers

When heated a sample consisting of only CaCO3 and MgCO3 yields a mixture of CaO and Mgo. If the weight of the combined oxides is

ExtremeBDS [4]

Answer:

38.83 % of CaCO3

61.17 % of MgCO3

Explanation:

where Moles of CaCO3 is equals to x and MgCO3 is y we have that...

CaCO3 molar mass = 100.09 g / mol = 100.09 x

MgCO3 molar mass = 84.31 g / mol = 84.31 y

decomposition reactions :

CaCO3 ---> CaO + CO2

MgCO3 ---> MgO + CO2

So we have that , Moles of CaO = Moles of CaCO3 = x

and Moles of MgO = Moles of MgCO3 = y

CaO molar mass = 56.08 g / mol

MgO molar mass = 40.30 g / mol

CaO = 56.08 x

MgO = 40.30 y

"If the weight of the combined oxides is equal to 51.00% of the initial sample weight,"

total mass of MgO and CaO = 51.00 % of Total Mass of MgCO3 and CaCO3

thus

56.08 x + 40.30 y = 0.51 ( 100.09 x + 84.31 y )

56.08 x + 40.30 y = 51.04 x + 42.99y

5.04 x = 2.7 y

y = 1.87 x

CaCO3 % in the sample

= 100.09 x × 100 / ( 100.09 x + 84.31 y )

= 10009 x / ( 100.09 x + 84.31 × 1.87 x )

= 10009 x / ( x ( 100.09 + 157.66 ) )

= 10009 / 257.75

= 38.83 %

MgCO3 % in the sample

= 100 - 38.83

= 61.17 %

7 0

3 years ago

a) Calculatethe molality, m, of an aqueous solution of 1.22 M sucrose, C12H22O11. The density of the solution is 1.12 g/mL.b) Wh

Contact [7]

Answer:

a) 1,74 molal

b) 37,2 %

c) 0,03

Explanation:

We are going to define sucrose as solute, water as solvent and the mix of both, the solution.

Let´s start with the data:

$Molarity = M = \frac{1,22 mol solute}{lts solution}$

We can assume as a calculus base, 1 liter of solution. So, in 1 liter of solution we have 1,22 moles of solute:

$1 lts solution * \frac{1,22 moles solute}{lts solution}=1,22 moles solute$

Knowing that the molality (m) is defined as mol of solute/kgs solvent, we have to calculate the mass of solvent on the solution. Remember our calculus base (1 lts of solution). In 1 lts of solution we have 1120 grams of solution.

$1 lts solution * \frac{1,12 grs solution}{mL solution}*\frac{1000 mL solution}{1 lts solution} = 1120 grs of solution$

With the molecular weight of solute (<em>Sum of: for carbon = 12*12=144; for hydrogen = 1*22=22 and for oxygen = 16*11=176. Final result = 342 grs per mol</em>), we can obtain the mass of solute:

$1,22 mol solute*\frac{342 grs solute}{1 mol solute} = 417,24 grs solute$

Now, the mass of solvent is: mass solvent = mass of solution - mass of solute. So, we have: 1120 - 417,24 = 702,76 grs of solvent = 0,70276 Kgs of solvent

$molality = m = \frac{1,22 mol solute}{0,70276 kgs solvent}= 1,74 molal$

For b) question we have that the mass percent of solute is hte ratio between the mass of solute and the mass of solution. So,

$%(w/w) = \frac{417,24 grs solute}{1120 grs solution} = 37,2%$

For c) question we have that the mole fraction of solute is the ratio between moles of solute and moles of solution. Let's calculate the moles of solution as follows: <em>Moles solution = moles solute + moles solvent.</em> First we have that the moles of solvent are (remember that the molecular weight of water for this calculus is 18 grs per mol):

$702,76 grs solvent*\frac{1 mol solvent}{18 grs solvent} = 39,04 moles solvent$

So, we have the moles of solution: 1,22 moles of solute + 39,04 moles of solvent = 40,26 moles of solution

Finally, we have:

$Mol frac solute = \frac{1,22 mol solute}{40,26 mol solution}= 0,03$

6 0

3 years ago

.War nickels, produced from 1942-1945 are composed of 56% copper, 35% silver and 9% manganese. How many moles of each element ar

AnnZ [28]

The moles of each element found in a 5.00 g nickel coin is calculated as below

moles =mass/molar mass

calculate the mass of each element =% composition of element/100 x total mass of nickel

Mn = 9/100 x5 = 0.45g
Cu=56/100 x5= 2.8 g
Ag= 35/100x5= 1.75 g
moles of each element is therefore=
Mn = 0.45g/54.94 = 8.19 x10^-3 moles

Ag=1.75g/107.87 g/mol = 0.0162 moles

Cu = 2.8 g/63.5 g/mol=0.0441 moles

3 0

3 years ago