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2 years ago

Q3. A student added water to calcium oxide to make calcium hydroxide

Chemistry

1 answer:

galina1969 [7]2 years ago

6 0

26.4 kg of calcium hydroxide (Ca(OH)₂)

Explanation:

We have the following chemical reaction:

CaO + H₂O → Ca(OH)₂

molecular weight of Ca(OH)₂ = atomic mass of Ca × number of Ca atoms + atomic mass of O × number of O atoms + atomic mass of H × number of H atoms

molecular weight of Ca(OH)₂ = 40 × 1 + 16 × 2 + 1 × 2 = 74 g/mole

molecular weight of CaO = atomic mass of Ca × number of Ca atoms + atomic mass of O × number of O atoms

molecular weight of CaO = 40 + 16 = 56 g/mole

number of moles = mass / molecular weight

number of moles of CaO = 20 / 56 = 0.357 kmoles

Taking in account the chemical reaction we devise the following reasoning:

if 1 kmole of CaO produces 1 kmole of Ca(OH)₂

then 0.357 kmoles of CaO produces X kmoles of Ca(OH)₂

X = (0.357 × 1) / 1 = 0.357 kmoles of Ca(OH)₂

mass = number of moles × molecular weight

mass of Ca(OH)₂ = 0.357 × 74 = 26.4 kg

Learn more about:

number of moles

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A 0.8870 g sample of a mixture of NaCl and KCl is dissolved in water, and the solution is then treated with an excess of AgNO3 t

bagirrra123 [75]

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.

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