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riadik2000 [5.3K]
3 years ago
7

luminum and oxygen react according to the following equation: 4Al(s) +3O2(g) --> 2Al2O3(s) What mass of Al2O3, in grams, can

be made by reacting 4.6 g Al with excess oxygen?
Chemistry
1 answer:
Slav-nsk [51]3 years ago
8 0

Answer: 8.7 grams

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 6.023\times 10^{23} of particles.  

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{4.6g}{27g/mol}=0.17moles

4Al(s)+3O_2(g)\rightarrow 2Al_2O_3(s)

As oxygen is in excess, Aluminium is the limiting reagent and limits the formation of products.

According to stoichiometry:

4 moles of aluminium give = 2 moles of Al_2O_3(s)

Thus 0.17 moles of aluminium give=\frac{2}{4}\times 0.17=0.085mol

Mass of Al_2O_3=moles\times {\text {molar mass}}=0.085\times 102g/mol=8.7g

Thus the mass of Al_2O_3(s)  is 8.7 grams

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Answer:

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation:

The formation reaction of CH_3OH will be,

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The intermediate balanced chemical reaction will be,

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CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l) , \Delta H_3=-726.4kJ/mole..[3]

Now we will reverse the reaction 3, multiply reaction 2 by 2  then adding all the equations, Using Hess's law:

We get :

C(graphite)+O_2(g)\rightarrow CO_2(g) , \Delta H_1=-393.5kJ/mole..[1]

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CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g) ,\Delta H_3=726.4kJ/mole [3]

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+2\times \Delta H_2+\Delta H_3

\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)

\Delta H=-238.7kJ/mole

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

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Hence, both Prilosec and Nexium are both equally effective.

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