Answer:
44.91% of Oxygen in Iron (III) hydroxide
Explanation:
To solve this question we must find the molar mass of Fe(OH)3 and the molar mass of the oxygen in this molecule. Percent composition will be:
<em>Molar mass Oxygen / molar mass Fe(OH)3 * 100</em>
<em />
<em>Molar mass Fe(OH)3 and oxygen:</em>
1Fe = 55.845g/mol*1 = 55.845
3O = 16.00g/mol*3 = 48.00 - Molar mass of Oxygen
3H = 1.008g/mol*3 = 3.024
55.845 + 48.00 + 3.024 =
106.869g/mol is molar mass of Fe(OH)3
% Composition of oxygen is:
48.00g/mol / 106.869g/mol * 100 =
<h3>44.91% of Oxygen in Iron (III) hydroxide</h3>
The balanced equation for the above neutralisation reaction is as follows;
Ca(OH)₂ + 2HCl ----> CaCl₂ + 2H₂O
Stoichiometry of Ca(OH)₂ to HCl is 1:2
number of Ca(OH)₂ moles reacted - 0.250 mol/L x 20.0 x 10⁻³ L = 5.00 x 10⁻³ mol
according to molar ratio of 1:2
number of HCl moles required = 2 x number of Ca(OH)₂ moles reacted
number of HCl moles = 5.00 x 10⁻³ x 2 = 10.0 x 10⁻³ mol
molarity of HCl solution - 0.250 M
there are 0.250 mol in volume of 1 L
therefore 10.0 x 10⁻³ mol in - 10.0 x 10⁻³ mol / 0.250 mol/L = 40.0 mL
40.0 mL of 0.250 M HCl is required
The matter will be consumed by other living organisms and the blood will settle to the bottom of the body
PE, GO, XY - I am probably wrong xoxoxoxoxxo
Answer:
moles of carbon dioxide produced are 410.9 mol.
Explanation:
Given data:
Mass of C₆H₁₄O₂ = 16.5 g
Moles of O₂ = 499 mol
Moles of CO₂ = ?
First of all we will write the balance chemical equation.
2C₆H₁₄O₂ + 17O₂ → 14CO₂ + 12H₂O
moles of C₆H₁₄O₂ = mass × molar mass
moles of C₆H₁₄O₂ = 16.5 g × 118 g/mol
moles of C₆H₁₄O₂ = 1947 mol
Now we compare the moles of CO₂ with moles of O₂ and C₆H₁₄O₂ from balance chemical equation.
O₂ : CO₂
17 : 14
499 : 14/17× 499 = 410.9 moles
C₆H₁₄O₂ : CO₂
2 : 14
1947 : 14/2× 1947 = 13629 moles
Oxygen will be limiting reactant so moles of carbon dioxide produced are 410.9 mol.
