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3 years ago

Please please help!

Chemistry

1 answer:

ratelena [41]3 years ago

3 0

The mass percent of potassium chloride is 1.386%

calculation

mass percent = actual mass/ Theoretical mass x 100

Actual mass = 9.35 g

Theoretical mass is calculated as below

Step 1 : write the equation for reaction

KCl + H₂O → KOH + HCl

Step 2: find the moles of H₂O

moles = mass÷ molar mass

The molar mass of H₂O = (2 x1 ) +(16) = 18 g/mol

moles is therefore = 162.98 g÷ 18 g/mol =9.054 moles

Step 3: use the mole ratio to determine the moles of KCl

KCl: H₂O is 1:1 therefore the moles of KCl is also = 9.054 moles

Step 4: find the theoretical mass of KCl

mass = moles x molar mass

from periodic table the molar mass of KCl = 39 +35.5 =74.5 g/mol

mass = 9.054 moles x 74.5 g/mol =674.5 g

Theoretical mass is therefore = 9.35 g/ 674.5 g x 100 = 1.386%

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A solution contains 42.0 g of heptane (C7H16) and 50.5 g of octane (C8H18) at 25 ∘C. The vapor pressures of pure heptane and pur

Kruka [31]

Answer:

(a) 22.3 torr; 5.6 torr; (b) 27.9 torr; (c) 77.7 % heptane; 23.3 % octane

(d) Heptane is more volatile than octane

Explanation:

We can use Raoult's Law to solve this problem.

It states that the partial pressure of each component of an ideal mixture of liquids is equal to the vapour pressure of the pure component multiplied by its mole fraction. In symbols,

$p_{i} = \chi_{i} p_{i}^{\circ}$

(a) Vapour pressure of each component

Let heptane be Component 1 and octane be Component 2.

(i) Moles of each component

$n_{1} = \text{42.0 g} \times \dfrac{\text{1 mol}}{\text{100.20 g}} = \text{0.4192 mol}\\n_{2} = \text{50.5 g} \times \dfrac{\text{1 mol}}{\text{114.23 g}} = \text{0.4421 mol}$

(ii) Total moles

$n_{\text{tot}} = 0.4192 + 0.4421 = \text{0.8613 mol}$

(iiii) Mole fractions of each component

$p_{1} = 0.4867 \times 45.8 = \textbf{22.3 torr}\\p_{2} = 0.5133 \times 10.9 = \ \textbf{5.6 torr}$

(iv) Partial vapour pressures of each component

$p_{1} = 0.4867 \times 45.8 = \textbf{22.3 torr}\\p_{2} = 0.5133 \times 10.9 = \textbf{5.6 torr}$

(b) Total pressure

$p_{\text{tot}} = p_{1} + p_{2} = 22.3 + 5.6 = \text{27.9 torr}$

(c) Mass percent of each component in vapour

$\chi_{1} = \dfrac{p_{1}}{p_{\text{Tot}}} = \dfrac{22.3}{27.9} =0.799\\\chi_{2} = \dfrac{p_{2}}{p_{\text{Tot}} }= \dfrac{5.6}{27. 9} =0.201$

The ratio of the mole fractions is the same as the ratio of the moles.

$\dfrac{n_{1}}{n_{2}} = \dfrac{0.799}{0.201}$

If we have 1 mol of vapour, we have 0.799 mol of heptane and 0.201 mol of octane

$m_{1} = 0.799 \times 100.20 = \text{80.1 g}\\m_{2} = 0.201\times 114.23 = \text{23.0 g}\\m_{\text{tot}} = 80.1 + 23.0 = \text{103.1 g}\\\\\text{ mass percent heptane} = \dfrac{80.1}{103.1} \times 100 \, \% = \mathbf{77.7\, \%}\\\\\text{ mass percent octane} = \dfrac{23.0}{103.1} \times 100 \, \% = \mathbf{22.3\, \%}}$

(d) Enrichment of vapour

The vapour is enriched in heptane because heptane is more volatile than octane.

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3 years ago

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agasfer [191]

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How many grams of hydrogen are needed to produce 1.80 g of water according to this equation? 2H2 + O2 → 2H2O

s344n2d4d5 [400]

Answer is c

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So the mass needed to produce 1.80g of water is
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3 years ago

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What is the percent composition of phosphoric acid H3PO4

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Hey there!:

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H = 1.00794 a.m.u

P = 30.973762 a.m.u

O = 15.9994 a.m.u

H % = [ ( 1.00794 * 3 ) / 97.9952 ] * 100

H% = 3.0857 %

P % = [ ( 30.973762 * 1 ) / 97.9952 ] * 100

P% = 31.6074 %

O % = [ ( 15.9994 * 4 ) / 97.9952 ] * 100

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