Question

Directions: Write the balanced equation for each of the following situations. \ In addition, list the reaction type. YOU MUST TELL THE AMOUNTS OF EVERY SUBSTANCE THAT REMAINS IN THE CONTAINER AT THE END OF THE REACTION. ASSUME THAT ALL REACTIONS GO TO COMPLETION. If only STOICHIOMETRY, tell how much of the excess reactant is used!!!! Reaction Type a. Combination Reaction b. Decomposition Reaction c. Single Displacement / THIS IS ONE TYPE OF Oxidation Reduction Reaction d. Precipitation Reaction e. Gaseous Reaction f. Neutralization Reaction g. Combustion Reaction 6. 8.924 g of copper(II) sulfate, pentahydrate is heated to drive off the water of hydration 6. Balanced Chemical Equation: At the completion of reactions: Grams of copper(II) sulfate pentahydrate: Grams of copper(II) sulfate: Grams of water:

Answer 1

The compound involved is copper(II) sulfate pentahydrate, that is, CuSO₄.5H₂O.

The heat will make the water separate, so the balanced equation is:

$CuSO_4.5H_2O(s)\to CuSO_4(s)+5H_2O(g)$

Since we start with one compound and end with more than on, this is a decomposition reaction.

Since we started with 8.924 g of CuSO₄.5H₂O, we can calculate how many moles are there to make the stoichiometry of the products.

The equation we use for transforming mass to number of moles or the other way around is:

$M=\frac{m}{n}_{}$

Where M is the molar weight of the compound.

The molar weight of CuSO₄.5H₂O is:

$\begin{gathered} M_{CuSO_4.5H_2O}=1\cdot M_{Cu}+1\cdot M_S+9\cdot M_O+10\cdot M_H \\ M_{CuSO_4.5H_2O}=(1\cdot63.546+1\cdot32.065+9\cdot15.9994+10\cdot1.00794)g/mol \\ M_{CuSO_4.5H_2O}=(63.546+32.065+143.9946+10.0794)g/mol \\ M_{CuSO_4.5H_2O}=249.6850g/mol \end{gathered}$

Thus, the number of moles is:

$\begin{gathered} M_{CuSO_{4}.5H_{2}O}=\frac{m_{CuSO_4.5H_2O}}{n_{CuSO_{4}.5H_{2}O}} \\ n_{CuSO_4.5H_2O}=\frac{m_{CuSO_4.5H_2O}}{M_{CuSO_{4}.5H_{2}O}}=\frac{8.924g}{249.6850g/mol}=0.035741\ldots mol \end{gathered}$

The reaction is 1 to 1 for CuSO₄ and 1 to 5 for H₂O, so this reaction will produce:

$\begin{gathered} n_{CuSO_4}=\frac{1}{1}n_{CuSO_4.5H_2O}=0.035741\ldots mol \\ n_{H_2O}=\frac{5}{1}n_{CuSO_4.5H_2O}=5\cdot0.035741mol=_{}0.17870\ldots mol \end{gathered}$

Now, we just need to convert these to mass, but first we need the molar weight of CuSO₄ and H₂O:

$\begin{gathered} M_{CuSO_4}=1\cdot M_{Cu}+1\cdot M_S+4\cdot M_O \\ M_{CuSO_4}=(1\cdot63.546+1\cdot32.065+4\cdot15.9994)g/mol \\ M_{CuSO_4}=(63.546+32.065+63.9976)g/mol \\ M_{CuSO_4}=159.6086g/mol \end{gathered}$$\begin{gathered} M_{H_2O}=2\cdot M_H+1\cdot M_O \\ M_{H_2O}=(2\cdot1.00794+1\cdot15.9994)g/mol \\ M_{H_2O}=(2.01588+15.9994)g/mol \\ M_{H_2O}=18.01528g/mol \end{gathered}$

So, we have:

$\begin{gathered} M_{CuSO_4}=\frac{m_{CuSO_4}}{n_{CuSO_{4}}} \\ m_{CuSO_4}=n_{CuSO_4}\cdot M_{CuSO_4}=0.035741\ldots mol\cdot159.6086g/mol\approx5.705g \end{gathered}$$\begin{gathered} M_{H_{2}O}=\frac{m_{H_2O}}{n_{H_{2}O}} \\ m_{H_2O}=n_{H_2O}\cdot M_{H_2O}=0.17870\ldots mol\cdot18.01528g/mol\approx3.219g \end{gathered}$

So, in the end we will have 0 g of copper(II) sulfate pentahydrate, approximately 5.705 g of copper(II) sulfate and approximately 3.219 g of water.