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hichkok12 [17]
3 years ago
12

48 g of magnesium sulfate are left after dehydration of magnesium sulfate heptahydrate . What was the original weight of the sam

ple before dehydration?
Chemistry
1 answer:
ValentinkaMS [17]3 years ago
7 0

Answer:

98.6g is the mass of the heptahydrate

Explanation:

The hydrated magnesium sulfate contains 7 moles of water per mole of magnesium sulfate. 1 mole of MgSO₄ dehydrated is 1 mole of MgSO₄.7H₂O

To solve this question we must find the moles of MgSO₄. These moles = Moles of MgSO₄.7H₂O. Then, we must convert the moles to mass:

<em>Moles MgSO₄ -Molar mass: 120.366g/mol-:</em>

48g * (1mol / 120.366g) = 0.40 moles MgSO₄ = moles MgSO₄.7H₂O

<em>Mass MgSO₄.7H₂O -Molar  mass: 246.47456 g/mol-:</em>

0.40 moles MgSO₄.7H₂O * (246.47456 g / 1mol) =

98.6g is the mass of the heptahydrate

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Answer:

\large \boxed{\text{ 0.066 mol/L}}

Explanation:

We are given the amounts of two reactants, so this is a limiting reactant problem.

1. Assemble all the data in one place, with molar masses above the formulas and other information below them.

Mᵣ:       58.44  

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m/g:     0.245

V/mL:                 50.

c/mmol·mL⁻¹:       0.0180

2. Calculate the moles of each reactant  

\text{Moles of NaCl} = \text{245 mg NaCl} \times \dfrac{\text{1 mmol NaCl}}{\text{58.44 mg NaCl}} = \text{4.192 mmol NaCl}\\\\\text{ Moles of AgNO}_{3}= \text{50. mL AgNO}_{3} \times \dfrac{\text{0.0180 mmol AgNO}_{3}}{\text{1 mL AgNO}_{3}} = \text{0.900 mmol AgNO}_{3}

3. Identify the limiting reactant  

Calculate the moles of AgCl we can obtain from each reactant.

From NaCl:  

The molar ratio of NaCl to AgCl is 1:1.

\text{Moles of AgCl} = \text{4.192 mmol NaCl} \times \dfrac{\text{1 mmol AgCl}}{\text{1 mmol NaCl}} = \text{4.192 mmol AgCl}

From AgNO₃:  

The molar ratio of AgNO₃ to AgCl is 1:1.  

\text{Moles of AgCl} = \text{0.900 mmol AgNO}_{3} \times \dfrac{\text{1 mmol AgCl}}{\text{1 mmol AgNO}_{3}} = \text{0.900 mmol AgCl}

AgNO₃ is the limiting reactant because it gives the smaller amount of AgCl.

4. Calculate the moles of excess reactant

                   Ag⁺(aq)  +  Cl⁻(aq) ⟶ AgCl(s)

 I/mmol:      0.900        4.192            0

C/mmol:    -0.900       -0.900        +0.900

E/mmol:      0                3.292          0.900

So, we end up with 50. mL of a solution containing 3.292 mmol of Cl⁻.

5. Calculate the concentration of Cl⁻

\text{[Cl$^{-}$] } = \dfrac{\text{3.292 mmol}}{\text{50. mL}} = \textbf{0.066 mol/L}\\\text{The concentration of chloride ion is $\large \boxed{\textbf{0.066 mol/L}}$}

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