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3 years ago

Please help...awarding brainliest if correct and lots of points.

Chemistry

1 answer:

abruzzese [7]3 years ago

8 0

Silver chloride produced : = 46.149 g

Limiting reagent : CuCl2

Excess remains := 3.74 g

<h3>Further explanation</h3>

Reaction

silver nitrate + copper(II) chloride ⇒ silver chloride + copper(II) nitrate

Required

silver chloride produced

limiting reagent

excess remains

Solution

Balanced equation

2AgNO3 (aq) + CuCl2 (s) → 2AgCl(s) + Cu(NO3)2(aq)

mol AgNO3 :

= 58.5 : 169,87 g/mol

= 0.344

mol CuCl2 :

=21.7 : 134,45 g/mol

= 0.161

mol ratio : coefficient of AgNO3 : CuCl2 :

= 0.344/2 : 0.161/1

= 0.172 : 0.161

CuCl2 as a limiting reagent

mol AgCl :

= 2/1 x 0.161

= 0.322

Mass AgCl :

= 0.322 x 143,32 g/mol

= 46.149 g

mol remains(unreacted) for AgNO3 :

= 0.344-(2/1 x 0.161)

= 0.022

mass AgNO3 remains :

= 0.022 x 169,87 g/mol

= 3.74 g

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<u>Answer:</u> The metal having molar mass equal to 26.95 g/mol is Aluminium

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

Molarity of NaOH solution = 0.5000 M

Volume of solution = 0.03340 L

Putting values in equation 1, we get:

$0.5000M=\frac{\text{Moles of NaOH}}{0.03340L}\\\\\text{Moles of NaOH}=(0.5000mol/L\times 0.03340L)=0.01670mol$

The chemical equation for the reaction of NaOH and sulfuric acid follows:

$2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, 0.01670 moles of NaOH will react with = $\frac{1}{2}\times 0.01670=0.00835mol$ of sulfuric acid

Excess moles of sulfuric acid = 0.00835 moles

Calculating the moles of sulfuric acid by using equation 1, we get:

Molarity of sulfuric acid solution = 0.5000 M

Volume of solution = 127.9 mL = 0.1279 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.5000M=\frac{\text{Moles of }H_2SO_4}{0.1279L}\\\\\text{Moles of }H_2SO_4=(0.5000mol/L\times 0.1279L)=0.06395mol$

Number of moles of sulfuric acid reacted = 0.06395 - 0.00835 = 0.0556 moles

The chemical equation for the reaction of metal (forming $M^{3+}$ ion) and sulfuric acid follows:

$2X+3H_2SO_4\rightarrow X_2(SO_4)_3+3H_2$

By Stoichiometry of the reaction:

3 moles of sulfuric acid reacts with 2 moles of metal

So, 0.0556 moles of sulfuric acid will react with = $\frac{2}{3}\times 0.0556=0.0371mol$ of metal

To calculate the molar mass of metal for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Mass of metal = 1.00 g

Moles of metal = 0.0371 moles

Putting values in above equation, we get:

$0.0371mol=\frac{1.00g}{\text{Molar mass of metal}}\\\\\text{Molar mass of metal}=\frac{1.00g}{0.0371mol}=26.95g/mol$

Hence, the metal having molar mass equal to 26.95 g/mol is Aluminium

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