Chemistry, 31.05.2020 03:00 cookiemonster0476

Hard water often contains dissolved Ca2+ and Mg2+ ions. One way to soften water is to add phosphates. The phosphate ion forms insoluble precipitates with calcium and magnesium ions, removing them from solution. Suppose that a solution is 0.054 M in calcium chloride and 0.093 M in magnesium nitrate. What mass of sodium phosphate would have to be added to 1.4 L of this solution to completely eliminate the hard water ions? Assume complete reaction. Enter a numerical answer only, in terms of grams to two significant figures.

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Answer from: mattmore0312

The mass of sodium phosphate required to completely eliminate ions from hard water has been 22.4269 g.

The hard water has been defined as the dissolved calcium and magnesium compounds in the solution. The hardness of water has been reduced by the elimination of these compounds from the sample.

The balanced reaction for the elimination of Calcium chloride and magnesium nitrate with sodium phosphate has been:

$\rm 3\;CaCl_2\;+\;2\;Na_3PO_4\;\rightarrow\;6\;NaCl\;+\;Ca_3(PO_4)_2$

$\rm 3\;Mg(NO_3)_2\;+\;2\;Na_3PO_4\;\rightarrow\;6\;NaNO_3\;+\;Mg_3(PO_4)_2$

Computation for the Mass of sodium phosphate

The moles of sample in the solution has been given by:

$\rm Moles=Molarity\;\times\;volume\;(L)$

The moles of calcium chloride in 0.054 M, 1.4 solution has been:

$\rm Moles\;CaCl_2=0.054\;\times\;1.4\\
Moles\;CaCl_2=0.0756\;mol$

The available moles of calcium chloride has been 0.0756 mol.

From the balanced chemical equation for the neutralization of 3 moles of Calcium chloride, 2 moles of sodium phosphate is required.

Thus, the moles of sodium phosphate required for the neutralization of calcium chloride has been:

$\rm 3\;mol\;CaCl_2=2\;mol\;Na_3PO_4\\
0.0756\;mol\;CaCl_2=\dfrac{2}{3}\;\times\;0.0756\;mol\;Na_2PO_4\\
 0.0756\;mol\;CaCl_2=0.05\;mol\;Na_3PO_4$

The moles of sodium phosphate required to neutralize calcium chloride has been 0.05 mol.

The moles of Magnesium nitrate in 0.093 M, 1.4 L solution has been:

$\rm Moles\;Mg(NO_3)_2=0.093\;\times\;1.4\\Moles\;CaCl_2=0.1302\;mol$

The available moles of Magnesium nitrate has been 0.1302 mol.

From the balanced chemical equation for the neutralization of 3 moles of Magnesium nitrate, 2 moles of sodium phosphate is required.

Thus, the moles of sodium phosphate required for the neutralization of Magnesium nitrate has been:

$\rm 3\;mol\;Mg(NO_3)_2=2\;mol\;Na_3PO_4\\0.1302\;mol\;CaCl_2=\dfrac{2}{3}\;\times\;0.1302\;mol\;Na_2PO_4\\ 0.1302\;mol\;CaCl_2=0.0868\;mol\;Na_3PO_4$

The moles of sodium phosphate required to neutralize Magnesium nitrate has been 0.0868 mol.

Thus, the total moles of sodium phosphate required has been:

$\rm Na_3PO_4=0.05+0.0868\;mol\\
Na_3PO_4=0.1368\;mol$

Thus, the total moles of sodium phosphate required has been 0.1368 mol.

The molar mass of sodium phosphate has been 163.94 g/mol.

The mass of 0.1368 mol Sodium phosphate has been:

$\rm Mass=Moles\;\times\;molar\;mass\\
Mass\;Na_3PO_4=0.1368\;\times\;163.94\;g\\
Mass\;Na_3PO_4=22.4269\;g$

Thus, the mass of sodium phosphate required to completely eliminate ions from hard water has been 22.4269 g.

Learn more about hard water, here:

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Answer from: poptropic7932

22.4269 grams of sodium phosphate must be added to 1.4 L of this solution to completely eliminate the hard water ions

Explanation:

We will first write the balanced equation for this scenario

3 CaCl2 + 2 Na3PO4 ----> 6 NaCl + Ca3 (PO4)2

3 Mg(NO3)2 + 2 Na3PO4 -----> 6 NaNO3 + Mg3 (PO4)2

The ratio here for both calcium chloride and magnesium nitrate is 3:2

The number of moles of each compound is equal to

$0.054 * 1.4 = 0.0756\\0.093* 1.4 = 0.1302$

Using the mole ratio of 3:2, convert each to moles of sodium phosphate.

0.0756 mole of CaCl2 is equal to $0.05\\$ Na3PO4

0.1302 mole of CaCl2 is equal to 0.0868 Na3PO4

Converting moles of sodium phosphate to grams of sodium phosphate we get

(0.05 +0.0868) * 163.94 g/mol

22.4269 grams of sodium phosphate must be added to 1.4 L of this solution to completely eliminate the hard water ions

Answer from: Quest

answer : (a) volume of the gas = 8.67 × $10^{-3}$ l

(b) second virial coefficient b = -1.21226 l/mole

solution : given, t = 300k

p = 20atm

z = 0.86

n = 8.2 mmoles = 8.2 × $10^{-3}$ moles

r = 0.082 l atm/k/mol

find, (a) volume of gas and (b) second virial coefficient b at 300k.

formula used : $z=\frac{pv}{nrt}$

and $z=1+\frac{b}{v}$

(a) $z=\frac{pv}{nrt}$

$v=\frac{znrt}{p}$

= $\frac{0.86\times 8.2\times[tex]10^{-3}$ \times 0.082\times 300}{20}[/tex]

= 8.67 × $10^{-3}$ l

(b) $z=1+\frac{b}{v}$

$b=(z-1)v$

= (0.00867-1) × 8.659

= - 8.5839 l/mole

Answer from: Quest

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Hard water often contains dissolved Ca2+ and Mg2+ ions. One way to soften water is to add phosphates...