Answer:
59.8%
Explanation:
First find the Mr of manganese (III) nitrate.
Mr of Mn(NO₃)₃ = 54.9 + (14 × 3) + (16 × 3 × 3) = <u>240.9</u>
Since we have to find the percentage composition of oxygen, we need to find the Mr of oxygen in the compound, which is:
Mr of (O₃)₃ = (16 × 3) × 3 = <u>144</u>
Now we can find percentage composition / percentage by mass of oxygen.
% composition = 
× 100
% composition = 
× 100 = <u>59.776%</u>
∴ % compostion of oxygen in maganese(III)nitrate is 59.8% (to 3 significant figures).
Answer:
Chemical reaction A, because the reactant is a compound
Explanation:
In a decomposition reaction, a compound is broken down into its components, so the number of products is greater than the number of reactants
Answer:
1.387 moles
Explanation:
Step 1:
The balanced equation for the reaction. This is illustrated below:
4Fe + 3O2 —> 2Fe2O3
Step 2:
Determination of the number of mole of Fe in 155.321g of Fe. This can be achieved by doing the following:
Mass of Fe = 155.321g
Molar Mass of Fe = 56g/mol
Number of mole of Fe =?
Number of mole = Mass/Molar Mass
Number of mole of Fe = 155.321/56
Number of mole of Fe = 2.774 mol
Step 3:
Determination of the number of mole of rust (Fe2O3) produced. This is illustrated below:
From the balanced equation above,
4 moles of Fe produced 2 moles of Fe2O3.
Therefore, 2.774 moles of Fe will produce = (2.774 x 2)/4 = 1.387 moles of Fe2O3.
Therefore, 1.387 moles of rust (Fe2O3) is produced from the reaction
I Think that the answer is 15.2096 Kilograms, but I might be wrong.
