Chemistry, 08.07.2019 14:30 christinapim

Suppose harry begins with the hydrate kal(so4)2·12h2o. after dehydration he finds that he is left with 3.0 g of the an-hydrate kal(so4)2. how many grams did he start with?

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Suppose harry begins with the hydrate kal(so4)2·12h2o. after dehydration he finds that he is left wi

Answer from: asauerha

Harry started with $\boxed{{\text{5}}{\text{.5 grams}}}$ of ${\text{KAl}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{2}}} \cdot{\text{12}}{{\text{H}}_{\text{2}}}{\text{O}}$ hydrated salt.

Further explanation:

Mole is the measure of the amount of substance. Mole is the relation between the mass of the substance and molar mass of a substance. It is defined as the mass of a substance in grams divided by its molar mass (g/mol).

The expression to calculate the number of moles is as follows:

${\text{Number of moles}}\;=\;\dfrac{{{\text{Given mass}}\left({\text{g}}\right)}}{{{\text{molar mass}}\left({{\text{g/mol}}}\right)}}$

Therefore, the formula to calculate the mass of the given compound is,

${\text{Mass}}\left({\text{g}} \right)=\left({{\text{Number of moles}}}\right)\left( {{\text{molar mass}}\left({{\text{g/mol}}}\right)}\right)$

The molar mass of ${\text{KAl}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{2}}} \cdot{\text{12}}{{\text{H}}_{\text{2}}}{\text{O}}$ is 474.3884g/mol.

The molar mass of ${\text{KAl}}{\left({{\text{S}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}$ is 258.2050 g/mol.

Therefore, mass of 12 ${{\text{H}}_2}{\text{O}}$ molecules is,

$\begin{aligned}{\text{Mass}}\left( {\text{g}} \right)&={\text{Moles}}\times{\text{molar mass of }{{\text{H}}_2}{\text{O}}\\&=12{\text{ mol }}\times 18{\text{ g/mol }}\\&=216{\text{ g}}\\\end{aligned}$

Since initially in ${\text{KAl}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{2}}} \cdot{\text{12}}{{\text{H}}_{\text{2}}}{\text{O}}$ , 1 mole of ${\text{KAl}}{\left({{\text{S}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}$ contained 12 moles of ${{\text{H}}_2}{\text{O}}$ . Thus 258.2050 g of ${\text{KAl}}{\left({{\text{S}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}$ should contains 216 g of ${{\text{H}}_2}{\text{O}}$ before evaporation.

Therefore, the mass of ${{\text{H}}_2}{\text{O}}$ contained by 1 g of ${\text{KAl}}{\left({{\text{S}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}$ is calculated as,

${\text{Mass}}\left({\text{g}}\right){\text{ of }}{{\text{H}}_2}{\text{O in 1 g of KAl}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{2}}}{\text{ }}=\dfrac{{{\text{216 g of }}{{\text{H}}_2}{\text{O}}}}{{258.2050{\text{ g of KAl}}{{\left({{\text{S}}{{\text{O}}_{\text{4}}}} \right)}_{\text{2}}}}}$

Therefore, the mass of ${{\text{H}}_2}{\text{O}}$ contained by 3.0 g of ${\text{KAl}}{\left({{\text{S}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}$ is calculated as follows:

$\begin{aligned}{\text{Mass}}\left({\text{g}} \right){\text{ of }}{{\text{H}}_2}{\text{O in 3 g of KAl}}{\left({{\text{S}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}&=\dfrac{{{\text{216 g of }}{{\text{H}}_2}{\text{O}}}}{{258.2050{\text{ g of KAl}}{{\left( {{\text{S}}{{\text{O}}_{\text{4}}}}\right)}_{\text{2}}}}}\times 3{\text{ g of KAl}}{\left({{\text{S}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}\\&=2.50{\text{ g }}\\\end{aligned}$

This 2.5 g is the mass of water that is evaporated from hydrated salt.

The total mass of hydrated of ${\text{KAl}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{2}}} \cdot{\text{12}}{{\text{H}}_{\text{2}}}{\text{O}}$ taken initially by harry is the sum of the mass of water that is evaporated and mass of dehydrated salt that is left after evaporation process.

$\begin{aligned}{\text{Total mass of hydrated salt}}&={{\text{m}}_{{\text{KAl}}{{\left( {{\text{S}}{{\text{O}}_{\text{4}}}}\right)}_{\text{2}}}}}+{{\text{m}}_{{{\text{H}}_2}{\text{O evaporated}}}}\\&=3.0{\text{ g}}+{\text{2}}{\text{.5 g}}\\&=5.5{\text{ g }}\\\end{aligned}$

Learn more:

1. Determine the number grams of solute in 500 ml of 0.189 M KOH.:

2. Rank the gases in decreasing order of effusion:

Answer details:

Grade: Senior school

Subject: Chemistry

Chapter: Mole concept

Keywords: harry, hydrate salt, KAl(SO4)2•12H2O, dehydration process, hydration process, anhydrated salt, KAl(SO4)2, 3.0 gram KAl(SO4)2, grams of start with, 5.5 g.

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Answer from: Quest

Mass is the amount of matter in an object. weight is the measurement of the pull of gravity on an object. say for example, you travel to the moon which has a different gravitational field compared to the gravitational field of earth. (gravitational fields: earth = 1, moon = 0.165) now when you are on the moon, the pull of gravity on you is different than on the earth. and as weight is the measurement of the pull of gravity on an object, that means yes, your weight can and will change on the moon. however, on the moon, on earth or even on jupiter, the amount of mass inside of you remains the same. meaning your physical mass remains constant. in conclusion it is possible for your weight to change as your mass remains constant.

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Suppose harry begins with the hydrate kal(so4)2·12h2o. after dehydration he finds that he is left wi...

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