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brilliants [131]
3 years ago
9

Copper(II) sulfate pentahydrate, CuSO4 ·5 H2O, (molar mass 250 g/mol) can be dehydrated by repeated heating in a crucible. Which

value is closest to the percentage mass of water lost from the total mass of salt in the crucible when the crucible undergoes repetitive heatings until a constant mass is reached?36%26%13%25%
Chemistry
1 answer:
prohojiy [21]3 years ago
8 0

Answer:

The water lost is 36% of the total mass of the hydrate

Explanation:

<u>Step 1:</u> Data given

Molar mass of CuSO4*5H2O = 250 g/mol

Molar mass of CuSO4 = 160 g/mol

<u>Step 2:</u> Calculate mass of water lost

Mass of water lost = 250 - 160 = 90 grams

<u>Step 3:</u> Calculate % water

% water = (mass water / total mass of hydrate)*100 %

% water = (90 grams / 250 grams )*100% = 36 %

We can control this by the following equation

The hydrate has 5 moles of H2O

5*18. = 90 grams

(90/250)*100% = 36%

(160/250)*100% = 64 %

The water lost is 36% of the total mass of the hydrate

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To calculate this, we will use the chemical equations as math equations and add them.

Firtly, we want the equation for the formation of CH₃CHO(g), so this will be the only product.

The reactants must be only the elements in their standard form, so C(g), O₂(g) and H₂(g). I would be more correct to use C(s), but since we odn't have information for this, we will assume it wants with C(g).

So, the reaction we want is:

C(g)+O_2(g)+H_2(g)\to CH_3CHO(g)

To balance the reaction, we can just do for eqach element separately, maintaining the coefficient of 1 on CH₃CHO(g):

\begin{gathered} 2C(g)+\frac{1}{2}O_2(g)+2H_2(g)\to CH_3CHO\mleft(g\mright) \\ \Delta H=? \end{gathered}

Now, we want to get to this equation adding the equations we want. We will apply the same operations to the enthalpies to get the enthalpy of formation.

The first given equation has the CH₃CHO(g), but it is on the left side and with coefficient of 2, so we need to invert the reaction and divided every coefficient by 2. The same operations have to be applied to the enthalpy, so the sign of the enthalpy will invert and it will be divided by 2:

\begin{gathered} 2CO_2(g)+2H_2O(l)\to CH_3CHO(g)+\frac{5}{2}O_2(g)_{} \\ \Delta H=\frac{2308.4kJ}{2}=1154.2kJ \end{gathered}

The second given equation has both C(g) and O₂(g), but since the third equation also has O₂(g), we will look just for C(g). We need 2 C(g), so we will need to doulbe the equation and its enthalpy:

\begin{gathered} 2C(g)+2O_2(g)\to2CO_2(g) \\ \Delta H=2\cdot-414.0kJ=-828.0kJ \end{gathered}

For the last, we will look into H₂(g) and since all the equations are balanced, O₂(g) will also be balanced by the end of it.

We need 2 H₂(g), so we don't need to do anything with this reaction:

\begin{gathered} 2H_2(g)+O_2(g)\to H_2O(l) \\ \Delta H=-597.4kJ \end{gathered}

Now, we add the equations:

\begin{gathered} \cancel{2CO_2\mleft(g\mright)}+\cancel{2H_2O\mleft(l\mright)}\to CH_3CHO(g)+\cancel{\frac{5}{2}O_2(g)}_{} \\ 2C(g)+\cancel{2O_2(g)}\to\cancel{2CO_2(g)} \\ 2H_2(g)+\cancel{O_2(g)}\to\cancel{H_2O(l)} \\ ------------------------------- \\ 2C(g)+\frac{1}{2}O_2(g)+2H_2(g)\to CH_3CHO(g) \end{gathered}

And we do the same with the enthalpies:

\begin{gathered} \Delta H=1154.2kJ+(-828.0kJ)+(-597.4kJ) \\ \Delta H=1154.2kJ-828.0kJ-597.4kJ \\ \Delta H=-271.2kJ \end{gathered}

This is the enthalpy for this reaction. To get the molar enthalpy of formation, we need to divide this value by the coefficient of CH₃CHO(g). Since this coefficient is 1, we have:

\Delta H_m=-\frac{271.2kJ}{1mol}=-271.2kJ\/mol

So, the molar enthalpy of formation given the data is -271.2 kJ/mol.

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