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svetlana [45]
3 years ago
14

Hard water often contains dissolved Ca2+ and Mg2+ ions. One way to soften water is to add phosphates. The phosphate ion forms in

soluble precipitates with calcium and magnesium ions, removing them from solution. Suppose that a solution is 5.5×10−2M in calcium chloride and 8.0×10−2M in magnesium nitrate.
What mass of sodium phosphate must be added to 1.5L of this solution to completely eliminate the hard water ions? Assume complete reaction.
Chemistry
1 answer:
avanturin [10]3 years ago
5 0
<span>5.5×10−2M in calcium chloride and 8.0×10−2M in magnesium nitrate.
What mass of sodium phosphate must be added to 1.5L of this solution to completely eliminate the hard water ion

1) Content of Ca (2+) ions

Calcium chloride = CaCl2

Ionization equation: CaCl2 ---> Ca (2+) + 2 Cl (-)

=> Molar ratios: 1 mol of CaCl2 : 1 mol Ca(2+) : 2 mol Cl(-)

Calculate the number of moles of CaCl2 in 1.5 liters of 5.5 * 10^-2 M solution

M = n / V => n = M*V = 5.5 * 10^ -2 M * 1.5 l = 0.0825 mol CaCl2

=> 0.0825 mol Ca(2+)

2) Number of phosphate ions needed to react with 0.0825 mol Ca(2+)

formula of phospahte ion: PO4 (3-)

molar ratio: 2PO4(3-) + 3Ca(2+) = Ca3 (PO4)2

Proportion: 2 mol PO4(3-) / 3 mol Ca(2+) = x / 0.0825 mol Ca(2+)

=> x = 0.0825 coml Ca(2+) * 2 mol PO4(3-) / 3 mol Ca(2+) = 0.055 mol PO4(3-)

3) Content of Mg(2+) ions

Ionization equation: Mg (NO3)2 ----> Mg(2+) + 2 NO3 (-)

Molar ratios: 1 mol Mg(NO3)2 : 1 mol Mg(2+) + 2 mol NO3(-)

number of moles of Mg(NO3)2 in 1.5 liter of 8.0 * 10^-2 M solution

n = M * V = 8.0 * 10^ -2 M * 1.5 liter = 0.12 moles Mg(NO3)2

ions of Mg(2+) = 0.12 mol Mg(NO3)2 * 1 mol Mg(2+) / mol Mg(NO3)2 = 0.12 mol Mg(2+)

4) Number of phosphate ions needed to react with 0.12 mol Mg(2+)

2PO4(3-) + 3Mg(2+) = Mg3(PO4)2

=> 2 mol PO4(3-) / 3 mol Mg(2+) = x / 0.12 mol Mg(2+)

=> x = 0.12 * 2/3 mol PO4(3-) = 0.16 mol PO4(3-)

5) Total number of moles of PO4(3-)

0.055 mol + 0.16 mol = 0.215 mol

6) Sodium phosphate

Sodium phosphate = Na3(PO4)

Na3PO4 ---> 3Na(+) + PO4(3-)

=> 1 mol Na3PO4 : 1 mol PO4(3-)

=> 0.215 mol PO4(3-) : 0.215 mol Na3PO4

mass in grams = number of moles * molar mass

molar mass of Na3 PO4 = 3*23 g/mol + 31 g/mol + 4*16 g/mol = 164 g/mol

=> mass in grams = 0.215 mol * 164 g/mol = 35.26 g

Answer: 35.26 g of sodium phosphate
</span>
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326L
dalvyx [7]

Answer:

Final number of moles = 0.675 mol

Mass = 2.7 g

Explanation:

Given data:

Initial volume of gas = 2.00 L

Final volume of gas = 2.70 L

Initial number of moles = 0.500 mol

Final number of moles = ?

Solution:

Formula:

V₁/n₁ = V₂/n₂

V₁ = Initial volume

n₁ = Initial number of moles

V₂ = Final volume of gas

n₂ = Final number of moles

Now we will put the values in formula.

2.00 L /0.500 mol = 2.70 L / n₂

n₂ = 2.70 L× 0.500 mol /2.00 L

n₂ = 1.35 L.mol / 2.00 L

n₂ = 0.675 mol

B)

Grams of helium added = ?

Solution:

Mass = number of moles × molar mass

Mass = 0.675 mol× 4 g/mol

Mass = 2.7 g

6 0
3 years ago
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ch4aika [34]
Answer is D earths distance from the sun
7 0
3 years ago
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gladu [14]

Answer:

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Explanation:

3 0
3 years ago
If you digested 25 mg of pure ferrocene in nitric acid and then diluted the resulting solution up to 500 mL with 1% aqueous hydr
Vilka [71]

Answer:

1.345*10⁻⁴ mol/L

15.023 mg/L

Explanation:

The chemical <u>formula of ferrocene</u> is Fe(C₅H₅)₂, thus its molecular weight is:

55.845 g/mol + 10*12g/mol + 10 *1g/mol = 185.845 g/mol

  • The moles of Fe contained in 25 mg (or 0.025 g) of ferrocene are:

0.025gFerrocene*\frac{1molFerrocene}{185.845g} *\frac{1mol Fe}{1molFerrocene}=1.345*10^{-4} molFe

  • The final volume is 500 mL, or 0.5 L. So the iron concentration in mol/L is:

\frac{1.345*10^{-4}molFe}{0.5L}= 2.69*10^{-4} mol/L

  • We can convert that value into mg/L:

2.69*10^{-4} \frac{molFe}{L} *\frac{55.845g}{1molFe}*\frac{1000mg}{1g}=15.023 mg/L

6 0
3 years ago
Consider the balanced chemical reaction below and determine the percent yield for iron(III) sulfide if 6.37 moles of iron(III) b
Sveta_85 [38]

Answer:

    Percent yield =  57.7 %

Explanation:

Given Data:

moles of iron(III) bromide = 6.37 mol

actual yield of iron(III) sulfide  = 1.84 mole

percent yield for iron(III) sulfide = ?

Reaction Given

                 2 FeBr₃ + 3 Na₂S ------------> Fe₂S₃+ 6 NaBr

Solution:

First we have to know the theoretical yield

We Know the Information given in the reaction

                         2 FeBr₃ + 3 Na₂S ------------> Fe₂S₃+ 6 NaBr

                         2 mol         3 mol                   1 mol

Now,

if two mole of iron(III) bromide (FeBr₃) give 1 mole of iron(III) sulfide (Fe₂S₃)

Then how many moles of Fe₂S₃ will be produced if 6.37 moles of FeBr₃ will be used

Apply the unity formula

                            2 mol  of   FeBr₃     ≅  1 mol of Fe₂S₃

                              6.37 mol  of FeBr₃ ≅  ? mol of Fe₂S₃

By doing cross multiplication

                          no. of mol of Fe₂S₃ = 1 mol x 6.37 mol/ 2 mol

                           no. of mol of Fe₂S₃ = 3.185 mol

So,

The theoretical yield is 3.19 mol

Formula Used to find Percent yield

         

             Percent yield = Actual yield/ theoretical yield x 100%

Now put all values in above equation

          Percent yield = 1.84 mol / 3.19 yield x 100 %

          Percent yield =  0.577x 100%

          Percent yield =  57.7 %

So the Percent yield of Iron(III) Sulfide =  57.7 %

6 0
3 years ago
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