300 g Al 90% purity react with H C L 36.5%.What is the total number of H atoms in final product?
This is a basic stoichiometric problem which needs to be broken down into five steps:
Step #1 : Write out and balance the chemical reaction.
2 Al + 6 HCl ---> 2 AlCl3 + 3 H2
Step # 2 : Determine the mass of pure Aluminum reacting
Al = 300g x .90% purity = 270 grams utilized
Step #3 : Utilize mass / mole stoichiometric ratio to find moles H2
Mass of known X
_____________________ = _________________
Atomic Weight X Coefficient Coefficient of Unknown
Moles H2 : 270 / (2 x 26.97) = X / 3 = 15.016 moles H2
Step 4: Find number of molecules H2 produced
15.016 moles x 6.023 x 10*23 = 9.044x 10*24 molecules H2
Step #5 : Find the number of atoms from molecules of H2
H2 is composed of 2 atoms
therefore, 2 x 9.044 x10*24 = 1.8088 x 10*25 atoms
The given stoichiometry problem can also be solved by simple unitary methods.
The purity level of aluminium is 90%. This means that
100 g impure metal is actually 90 g Al.
`rArr` 300 g impure metal is actually 90*300/100 =270 g Al.
Now, the balanced chemical equation for the reaction of Al with HCl is:
2Al(s) + 6HCl(aq.) `rarr` 2AlCl3(aq.) + 3H2(g)
Assuming 36.5% HCl is present in sufficient amounts, Al should be the limiting reagent. Then, it follows from the stoichiometry of the reaction that:
2*26.98 g Al react to produce 3*2*6.022*10^23 number of hydrogen atoms.
`rArr` 270 g Al should react to produce `((3*2*6.02205*10^23)*270)/(2*26.98)`
`= 1.81*10^25`number of hydrogen atoms.
Therefore, total number of hydrogen atoms in the final product is `1.81*10^25` .
You need to write the chemical reaction of Al and hydrochloric acid, such that:
`Al + HCl -> AlCl_3 + H_2`
You need to balance the equation, such that:
`2Al + 6HCl -> 2AlCl_3 + 3H_2`
You need to evaluate the amount of pure aluminium from the total amount of `300g Al` of` 90%` purity, such that:
`m_Al = (90/100)*300 => m_Al = 270 g`
You need to evaluate the number of moles of H, such that:
`nu_(H_2) = (3*270g)/2*27 = 15 ` moles
You need to find the number of atoms in 15 moles of H, such that:
`A_1 = 15*2*6.023*10^23 => A_1 = 180.69*10^23` atoms of hydrogen
You also can find atoms of hydrogen in water resulted from `36.5%HCl`
You need to find how many grams of `36.5%HCl` solution are, knowing that 1 mol of `HCl` has `36.5 g` , such that:
`m_(HCl) = (270*6*36.5)/(2*27) = 1095HCl`
You need to evaluate the mass of solution of `36.5%HCl` such that:
`m_s = (1095*100)/36.5 => m_s = 3000g` solution of `36.5%HCl`
You may evaluate now the mass of water resulted from `36.5%HCl` , such that:
`m_(H_2O) = 3000 - 1095 = 1905`
You need to evaluate the number of atoms of H in water resulted from `36.5%HCl,` such that:
`A_2 = (1905*2*6.023*10^23)/18 => A_2 = 1247.87*10^23 ` atoms of H
You need to evaluate the total number of hydrogen atoms, such that:
`N_(H) = A_1 + A_2 = 180.69*10^23 + 1247.87*10^23 = 1455.56*10^23`
Hence, evaluating the total number of hydrogen atoms, under the given conditions, yields `N_(H) = 1455.56*10^23.`