Answer:
Solution Density of aluminium = 2.7 g/Cm 3 In kg/ m 3 = 27 × 1000 10 =2700 kg/ m 3
Explanation:
Not much of one
Since you have not included the chemical reaction I will explain you in detail.
1) To determine the limiting agent you need two things:
- the balanced chemical equation
- the amount of every reactant involved as per the chemical equation
2) The work is:
- state the mole ratios of all the reactants: these are the ratios of the coefficientes of the reactans in the balanced chemical equation.
- determine the number of moles of each reactant with this formula:
number of moles = (mass in grams) / (molar mass)
- set the proportion with the two ratios (theoretical moles and actual moles)
- compare which reactant is below than the stated by the theoretical ratio.
3) Example: determine the limiting agent in this reaction if there are 100 grams of each reactant:
i) Chemical equation: H₂ + O₂ → H₂O
ii) Balanced chemical equation: 2H₂ + O₂ → 2H₂O
iii) Theoretical mole ration of the reactants: 2 moles H₂ : 1 mol O₂
iv) Covert 100 g of H₂ into number of moles
n = 100g / 2g/mol = 50 mol of H₂
v) Convert 100 g of O₂ to moles:
n = 100 g / 32 g/mol = 3.125 mol
vi) Actual ratio: 50 mol H₂ / 3.125 mol O₂
vii) Compare the two ratios:
2 mol H₂ / 1 mol O ₂ < 50 mol H₂ / 3.125 mol O₂
Conclusion: the actual ratio of H₂ to O₂ is greater than the theoretical ratio, meaning that the H₂ is in excess respect to the O₂. And that means that O₂ will be consumed completely while some H₂ will remain without react.
Therefore, the O₂ is the limiting reactant in this example.
By 1.23 x 1024 you mean 10 to the power of 24 molecules? If so all you need to do is divide the number of molecules you have by Avagadros number, 6.022 x 10^23. This will give you the mols of water, or the mols of anything, since there is always 6.022 x 10^23 molecules in 1 mol of substance.
1.23x10^24 atoms/6.022x10^23 atom/mol = 2.04 mol H20
Answer:
- <u><em>1.12 liters</em></u>
Explanation:
<u>Calculating number of moles</u>
- Molar mass of O₂ = 32 g
- n = Given weight / Molar mass
- n = 1.6/32
- n = 0.05 moles
<u>At STP</u>
- One mole of O₂ occupies 22.4 L
- Therefore, 0.05 moles will occupy :
- 22.4 L x 0.05 = <u><em>1.12 L</em></u>
