Answer:
163.2g
Explanation:
First let us generate a balanced equation for the reaction. This is shown below:
4Al + 3O2 —> 2Al2O3
From the question given, were were told that 3.2moles of aluminium was exposed to 2.7moles of oxygen. Judging by this, oxygen is excess.
From the equation,
4moles of Al produced 2moles of Al2O3.
Therefore, 3.2moles of Al will produce = (3.2x2)/4 = 1.6mol of Al2O3.
Now, let us covert 1.6mol of Al2O3 to obtain the theoretical yield. This is illustrated below:
Mole of Al2O3 = 1.6mole
Molar Mass of Al2O3 = (27x2) + (16x3) = 54 + 48 =102g/mol
Mass of Al2O3 =?
Number of mole = Mass /Molar Mass
Mass = number of mole x molar Mass
Mass of Al2O3 = 1.6 x 102 = 163.2g
Therefore the theoretical of Al2O3 is 163.2g
An oxygen atom is 16x more massive than a hydrogen atom.
You can figure this out by comparing the atomic masses of the two elements: oxygen has an atomic mass number of 16, and hydrogen has an atomic mass number of 1. Thus, an oxygen atom is 16 times more massive than a hydrogen atom.
Answer:
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Answer:
83.64%.
Explanation:
∵ The percent yield = (actual yield/theoretical yield)*100.
actual yield of CO₂ = 2300 g.
- We need to find the theoretical yield of CO₂:
For the reaction:
<em>CH₄ + 2O₂ → 2H₂O + CO₂,</em>
1.0 mol of CH₄ react with 2 mol of O₂ to produce 2 mol of H₂O and 1.0 mol of CO₂.
- Firstly, we need to calculate the no. of moles of 1000 g of CH₄ using the relation:
<em>no. of moles of CH₄ = mass/molar mass</em> = (1000 g)/(16.0 g/mol) = <em>62.5 mol.</em>
<u><em>Using cross-multiplication:</em></u>
1.0 mol of CH₄ produces → 1.0 mol of CO₂, from stichiometry.
∴ 62.5 mol of CH₄ produces → 62.5 mol of CO₂.
- We can calculate the theoretical yield of carbon dioxide gas using the relation:
∴ The theoretical yield of CO₂ gas = n*molar mass = (62.5 mol)(44.0 g/mol) = 2750 g.
<em>∵ The percent yield = (actual yield/theoretical yield)*100.</em>
actual yield = 2300 g, theoretical yield = 2750 g.
<em>∴ the percent yield</em> = (2300 g/2750 g)*100 = <em>83.64%.</em>
Answer:
b
Explanation:
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