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sveticcg [70]
3 years ago
7

Using the following thermochemical data, what is the change in enthalpy for the following reaction: 3H2(g) + 2C(s) + ½O2(g) C2H5

OH(l) , ΔH = –1367 kJ/mol , ½ ,
Chemistry
1 answer:
Tamiku [17]3 years ago
6 0
Given: 

<span>3H2(g) + 2C(s) + ½O2(g)     </span>→   <span>C2H5OH(l) , ΔH = –1367 kJ/mol 

Change is enthalpy is mathematically expressed as enthalpy of product minus enthalpy of reactant. If value of enthalpy change is negative, it suggest reaction is exothermic. If enthalpy change is positive, it suggest reaction is endothermic.

In present case, enthalpy change is -1367 kJ/mol.</span>
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Which of the following compounds will be most soluble in ethanol (CH3CH2OH)?hexane (CH3CH2CH2CH2CH2CH3)ethylene glycol (HOCH2CH2
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Magnesium (used in the manufacture of light alloys) reacts with iron(III) chloride to form magnesium chloride and iron. A mixtur
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<u>Answer:</u> The limiting reactant is magnesium and mass of excess reactant present in the vessel is 96.35 grams.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For magnesium:</u>

Given mass of magnesium = 41.0 g

Molar mass of magnesium = 24 g/mol

Putting values in equation 1, we get:

\text{Moles of magnesium}=\frac{41.0g}{24g/mol}=1.708mol

  • <u>For iron(III) chloride:</u>

Given mass of iron(III) chloride = 175.0 g

Molar mass of iron(III) chloride = 162.2 g/mol

Putting values in equation 1, we get:

\text{Moles of iron(III) chloride}=\frac{175g}{162.2g/mol}=1.708mol

The chemical equation for the reaction of magnesium and iron(III) chloride follows:

3Mg+2FeCl_3\rightarrow 3MgCl_2+2Fe

By Stoichiometry of the reaction:

3 moles of magnesium reacts with 2 moles of iron(III) chloride

So, 1.708 moles of magnesium will react with = \frac{2}{3}\times 1.708=1.114mol of iron(III) chloride

As, given amount of iron(III) chloride is more than the required amount. So, it is considered as an excess reagent.

Thus, magnesium is considered as a limiting reagent because it limits the formation of product.

Moles of excess reactant left (iron(III) chloride) = [1.708 - 1.114] = 0.594 moles

Now, calculating the mass of iron(III) chloride from equation 1, we get:

Molar mass of iron(III) chloride = 162.2 g/mol

Moles of iron(III) chloride = 0.594 moles

Putting values in equation 1, we get:

0.594mol=\frac{\text{Mass of iron(III) chloride}}{162.2g/mol}\\\\\text{Mass of iron(III) chloride}=(0.594mol\times 162.2g/mol)=96.35g

Hence, the limiting reactant is magnesium and mass of excess reactant present in the vessel is 96.35 grams.

6 0
3 years ago
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