1) Write the balanced chemical equation:
2Al + 6HCl ---> 2AlCl3 + 3H2
2) Calculate the molar masses of Al and HCl and use them to convert the data to moles
Al: 27g/mol => 10.7 gAl / 27g/mol = 0.396 mol Al
HCl: 1g/mol + 35.5 g/mol = 36.5g/mol => 42.5gHCl / 36.5g/mol = 1.164 mol HCl
Theoretical ratio: 6 mol HCl / 2 mol Al = 3:1
Actual ratio: 1.164 mol HCl / 0.396 mol Al = 2.94 : 1
Then there is a little bit less HCl than the predicted by theoretical ratio, which means that this is the limitant reagent.
3) Use the amount of HCl to make the calculations of the proucts obtained.
Theoretical ratio: 3 mol H2 / 6 mol HCl = 1:3
1.164 mol HCl * 1 mol H2 / 3 mol HCl = 0.388 mol H2
4) Use the ideal gas formula to obtain the volume
pV = n RT
p = 725 mmHg * 1atm / 760mmHg = 0.954 atm
n = 0.388 mol
R = 0.082 atm*liter /K*mol
T = 47 °C + 273.15 = 320.15K
V = nRT/p = 0.388 mol * 0.082 [atm*liter/K*mol] * 320.15K / 0.954 atm = 10.7 liter
Answer: 10.7 liter
Answer:
121.5 L will be the new volume
Explanation:
To solve this we can use the Ideal Gases Law
P . V = n . R . T for the two situations. But the moles of gas in the balloon are always the same so:
P . V / R .T = P . V / R. T
R is a universal constant, we can reject it. So:
P₁ . V₁ / T₁ = P₂ . V₂ / T₂
We also need to convert T° C to T°K → T°C + 273
30°C + 273 = 303 K and -48°C + 273 = 225K
Let's replace → (50.4L . 100 kPa) / 303K = (30.8 kPa . V₂) / 225K
((50.4L . 100 kPa) / 303K ) . 225K = 30.8 kPa . V₂
3742.5 kPa . L = 30.8 kPa . V₂
3742.5 kPa . L / 30.8 kPa = V₂ → 121.5 L
Answer:
b. temperature difference
The product of a reaction between these two elements is .
Explanation:
The oxidation state of an ion in a compound is equal to its charge.
The aluminum having a charge of +3 because oxidation state is +3
The oxide is having charge of -2
The product of these reactants will produce a chemical compound.
The compound formed is i.e Aluminium oxide. The compound while getting formed will share the charge and cation A+ will have the charge of anion and anion will have the charge of cation. This will result in a compound as there should be a neutral charge on the compound formed.
The <em>+</em><em>3 charge of the cation Al+ will go to anion oxide O2- and the charge of anion -2 will go with cation Al+. </em>
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