Chemistry, 29.07.2019 20:20 nicolascorrea0207

A0.8870 g sample of a mixture of nacl and kcl is dissolved in water, and the solution is then treated with an excess of agno3 to yield 2.142 g of agcl. calculate the percent by mass of each compound in the mixture

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Answer from: alyssaflores

Answer : The percent by mass of NaCl and KCl are, 92.22 % and 7.78 % respectively.

Explanation :

As we know that when a mixture of NaCl and KCl react with excess AgNO_3 then the silver ion react with the chloride ion in both NaCl and KCl to form silver chloride.

Let the mass of NaCl be, 'x' grams and the mass of KCl will be, (0.8870 - x) grams.

The molar mass of NaCl and KCl are, 58.5 and 74.5 g/mole respectively.

First we have to calculate the moles of NaCl and KCl.

$\text{Moles of }NaCl=\frac{\text{Mass of }NaCl}{\text{Molar mass of }NaCl}=\frac{xg}{58.5g/mole}=\frac{x}{58.5}moles$

$\text{Moles of }KCl=\frac{\text{Mass of }KCl}{\text{Molar mass of }KCl}=\frac{(0.8870-x)g}{74.5g/mole}=\frac{(0.8870-x)}{74.5}moles$

As, each mole of NaCl and KCl gives one mole of chloride ions.

So, moles of chloride ions in NaCl = $\frac{x}{58.5}moles$

Moles of chloride ions in KCl = $\frac{(0.8870-x)}{74.5}moles$

The total moles of chloride ions = $\frac{x}{58.5}moles+\frac{(0.8870-x)}{74.5}moles$

Now we have to calculate the moles of AgCl.

As we know that, this amount of chloride ion is same as the amount chloride ion present in the AgCl precipitate. That means,

Moles of AgCl = Moles of chloride ion = $\frac{x}{58.5}moles+\frac{(0.8870-x)}{74.5}moles$

Now we have to calculate the moles of AgCl.

The molar mass of AgCl = 143.32 g/mole

$\text{Moles of }AgCl=\frac{\text{Mass of }AgCl}{\text{Molar mass of }AgCl}=\frac{2.142g}{143.32g/mole}=0.0149moles$

Now we have to determine the value of 'x'.

Moles of AgCl = $\frac{x}{58.5}moles+\frac{(0.8870-x)}{74.5}moles$

0.0149 mole = $\frac{x}{58.5}moles+\frac{(0.8870-x)}{74.5}moles$

By solving the term, we get the value of 'x'.

x=0.818g

The mass of NaCl = x = 0.818 g

The mass of KCl = (0.8870 - x) = 0.8870 - 0.818 = 0.069 g

Now we have to calculate the mass percent of NaCl and KCl.

$\text{Mass percent of }NaCl=\frac{\text{Mass of }NaCl}{\text{Total mass of mixture}}\times 100=\frac{0.818g}{0.8870g}\times 100=92.22\%$

$\text{Mass percent of }KCl=\frac{\text{Mass of }KCl}{\text{Total mass of mixture}}\times 100=\frac{0.069g}{0.8870g}\times 100=7.78\%$

Therefore, the percent by mass of NaCl and KCl are, 92.22 % and 7.78 % respectively.

Answer from: Quest

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is $1s^{2}2s^{2}2p^{6}3s^{1}$ and electronic configuration of phosphorous is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}$ .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

$3na + p \rightarrow na_{3}p$

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.

Electrons are transferred from atoms of to atoms of this transfer makes the sodium atoms and the

Answer from: Quest

ill try to . no promises tho.

explanation:

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