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3 years ago

Molar mass of KAl(SO4)^2·12H2O

Chemistry

2 answers:

xeze [42]3 years ago

3 0

The molar mass is 477.09 g/mol.

To calculate the molar mass, let’s rewrite the formula as KAlS₂O₈H₂₄O₁₂.Then we can write:

1 K = 1 × 39.10 u = 39.10 u

1 Al = 1 × 26.98 u = 29.68 u

2 S = 2 × 32.06 u = 64.12 u

8 O = 8 × 16.00 u = 128.00 u

24 H = 24 × 1.008 u = 24.192 u

12 O = 12 × 16.00 u = 192.00 u

TOTAL = 477.09 u

The molar mass is 477.09 g/mol.

Vika [28.1K]3 years ago

3 0

Answer: The molar mass of $KAl(SO_4)_2.12H_2O$ is 474.23 g/mol

Explanation:

Molar mass is defined as the sum of the mass of all the atoms each multiplied its atomic masses that are present in the molecular formula of a compound. It is expressed in g/mol.

We know that:

Molar mass of potassium atom = 39.09 g/mol

Molar mass of aluminium atom = 26.98 g/mol

Molar mass of sulfur atom = 32.06 g/mol

Molar mass of oxygen atom = 15.99 g/mol

Molar mass of hydrogen atom = 1.01 g/mol

The chemical formula of the given compound is $KAl(SO_4)_2.12H_2O$

Molar mass of $KAl(SO_4)_2.12H_2O=[(1\times 39.09)+(1\times 26.98)+2(32.06+(4\times 15.99))+12((2\times 1.01)+15.99)]=474.23g/mol$

Hence, the molar mass of $KAl(SO_4)_2.12H_2O$ is 474.23 g/mol

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The equilibrium constant for the reaction 2x(g)+y(g)=2z(g) is 2.25 . what would be the concentration of y at equilibrium with 2

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$[\text{Y}] \approx0.337\;\text{mol}\cdot\text{dm}^{-3}$ at equilibrium.

<h3>Explanation</h3>

Concentration for each of the species:

$[\text{X}] = \dfrac{n}{V} = 2\; \text{mol}\cdot \text{dm}^{-3}$ ;
$[\text{Y}] = \dfrac{n}{V} = 0\; \text{mol}\cdot \text{dm}^{-3}$ ;
$[\text{Z}] = \dfrac{n}{V} = 3\; \text{mol}\cdot \text{dm}^{-3}$ .

There was no Y to start with; its concentration could only have increased. Let the change in $[\text{Y}]$ be $+x \; \text{mol}\cdot \text{dm}^{-3}$ .

Make a $\textbf{RICE}$ table.

Two moles of X will be produced and two moles of Z consumed for every one mole of Y produced. As a result, the change in $[\text{X}]$ will be $+2\;x \; \text{mol}\cdot \text{dm}^{-3}$ and the change in $[\text{Z}]$ will be $-2\;x \; \text{mol}\cdot \text{dm}^{-3}$ .

$\begin{array}{l|ccccc}\textbf{R}\text{eaction}&2\; \text{X}\; (g) & + &\text{Y}\; (g) & \rightleftharpoons &2 \; \text{Z}\; (g)\\\textbf{I}\text{nitial Condition}\; (\text{mol}\cdot\text{dm}^{-3})& 2 & &0 & & 3 \\\textbf{C}\text{hange in Concentration}\; (\text{mol}\cdot\text{dm}^{-3})\;& +2\;x & &+x &&-2\;x\\\textbf{E}\text{quilibrium Condition}\; (\text{mol}\cdot\text{dm}^{-3})& & &&&\end{array}$ .

Add the value in the C row to the I row:

What's the equation of $K_c$ for this reaction? Raise the concentration of each species to its coefficient. Products go to the numerator and reactants are on the denominator.

$K_c = \dfrac{[\text{Z}]^{2}}{[\text{X}]^{2} \cdot[\text{Y}]}$ .

$K_c = 2.25$ . As a result,

$\dfrac{[\text{Z}]^{2}}{[\text{X}]^{2} \cdot[\text{Y}]} = \dfrac{(3-2x)^{2}}{(2+2x)^{2} \cdot x} = K_c = 2.25$ .

$(3-2\;x)^{2}= 2.25 \cdot(2+2\;x)^{2} \cdot x\\4\;x^{2} - 12 \;x + 9 = 2.25 \;(4\;x^{3} + 8 \;x^{2} + 4 \;x)\\4\;x^{2} - 12\;x + 9 = 9 \;x^{3} + 18\;x^{2} + 9\;x\\9\;x^{3} + 14\;x^{2} + 21\;x - 9 = 0$ .

The degree of this polynomial is three. Plot the equation $y = 9\;x^{3} + 14\;x^{2} + 21\;x - 9$ on a graph and look for any zeros. There's only one zero at $x \approx 0.337$ . All three concentrations end up greater than zero.

Hence the equilibrium concentration of Y: $0.337\;\text{mol}\cdot\text{dm}^{-3}$ .

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3 years ago