Aluminium reacts with dilute sulfuric acid based on the following reaction:
<span>2Al + 3H2SO4 ..............> Al2 (SO4)3 + 3H2
From the periodic table:
mass of aluminium = 27 grams
mass of hydrogen = 1 gram
mass of oxygen = 16 grams
mass of sulfur = 32 grams
Therefore:
molar mass of aluminium = 27 grams
molar mass of sulfuric acid = 2(1) + 32 + 4(16) = 98 grams
From the balanced chemical equation:
2 moles of aluminium react with 3 moles of dilute sulfuric acid.
This means that 34 grams of Al react with 294 grams of the acid
To get the amount of aluminium that reacts with </span><span>5.890 g of sulfuric acid, we will do cross multiplication as follows:
</span>amount of Al = (<span>5.890 x 34) / 294 = 0.6811 grams</span>
There is a missing portion of this question which shows the reaction that needs balancing:
"In a balanced equation, the same number of each kind of atom is shown on each side of the equation. Calculate the number of iron (Fe), oxygen (O), and carbon atoms (C).
Fe2O3+ 3CO --> 2Fe + 3CO<span>2
</span><span>Based on these values, is the equation balanced?</span><span>"
</span>
To check if this equation is balanced we simply compare the number of each element on each side of the equation.
On the reactant side of the equation we have:
2 Fe atoms
6 O atoms
3 C atoms
On the product side of the equation we have:
2 Fe atoms
6 O atoms
3 C atoms
Therefore, both side of the reaction have the correct and equal number of each atom, so the equation is balanced.
Answer:
Explanation:
For a first order reaction the rate law is:
![v=\frac{-d[A]}{[A]}=k[A]](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B-d%5BA%5D%7D%7B%5BA%5D%7D%3Dk%5BA%5D)
Integranting both sides of the equation we get:
![\int\limits^a_b {\frac{d[A]}{[A]}} \, dx =-k\int\limits^t_0 {} \, dt](https://tex.z-dn.net/?f=%5Cint%5Climits%5Ea_b%20%7B%5Cfrac%7Bd%5BA%5D%7D%7B%5BA%5D%7D%7D%20%5C%2C%20dx%20%3D-k%5Cint%5Climits%5Et_0%20%7B%7D%20%5C%2C%20dt)
where "a" stands for [A] (molar concentration of a given reagent) and "b" is {A]0 (initial molar concentration of a given reagent), "t" is the time in seconds.
From that integral we get the integrated rate law:
![ln\frac{[A]}{[A]_{0} } =-kt](https://tex.z-dn.net/?f=ln%5Cfrac%7B%5BA%5D%7D%7B%5BA%5D_%7B0%7D%20%7D%20%3D-kt)
![[A]=[A]_{0}e^{-kt}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA%5D_%7B0%7De%5E%7B-kt%7D)
![ln[A]=ln[A]_{0} -kt](https://tex.z-dn.net/?f=ln%5BA%5D%3Dln%5BA%5D_%7B0%7D%20-kt)
![k=\frac{ln[A]_{0}-ln[A]}{t}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7Bln%5BA%5D_%7B0%7D-ln%5BA%5D%7D%7Bt%7D)
therefore k is
If one were to match the ratio of atoms of the elements found in this molecular formula of artificial sweetener it would be :
Carbon - 7 atoms
Hydrogen - 5 atoms
Nitrogen - 1 atom
Oxygen - 3 atoms.
