Answer:
Cancel out CO because it appears as a reactant in one intermediate reaction and a product in the other intermediate reaction.
Explanation:
The CO appears twice hence in he intermediate reaction it only forms path of the enabling reagents and it further reacts to form the final product. Accounting for the CO in the intermediate reaction that undergoes further reaction will impact on the stoichiometry of the reaction.
The oxidation state of the elements in the compounds are:
CoH₂:
FeBr₃:
<h3>What is the oxidation states of the elements in the given compounds?</h3>
The oxidation states of the elements in each of the given compounds is determined as follows:
Cobalt dihydride, CoH₂
Co = +2
H = -1
Iron (iii) bromide, FeBr₃
Fe = +3
Br = -1
In conclusion, the oxidation state of the elements are charges they have in the compound.
Learn more about oxidation state at: brainly.com/question/27239694
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Answer:
First confirm the reaction is balanced:
C3H8 + 5O2 --> 3CO2 + 4H20 (3 cabon - check; 8 hydrogen - check; 10 oxygen - check).
a) In the equation there is a 5:1 ratio between propane and oxygen. We also know that number of mole is proportional to pressure and volume. Since pressure is constant (STP) then the volume of O2 is 7.2 * 5 = 36 litres.
b) For a near ideal gas that PV = nRT (combined gas law). So for 7.2 litres propane we find n(propane) = 101.3 * 7.2/8.314*298 ~ 0.29 mole (using metric units throughout for simplicity).
There is a 1:3 ratio between propane and CO2. Therefore 3 * 0.29 = 0.87 mole of CO2 is produced.
MW(CO2) ~ 44 g/mol. Therefore m(CO2) = 44 * 0.87 ~ 38.3 g
c) We know we need more oxygen than propane (due to the 1:5 ratio) so oxygen is the limiting reagent. Again Volume is proportional to number of mole and we see there is a 5:4 ratio between oxygen and water. Therefore the volume of water vapour produced will be (4/5) * 15 = 12 litres.
The other questions use the same technique and will give you some much needed practice.
Explanation:
The given reaction is:
3Fe + 4H2O → Fe3O4 + 4H2
Given:
Mass of Fe = 354 g
Mass of H2O = 839 g
Calculation:
Step 1 : Find the limiting reagent
Molar mass of Fe = 56 g/mol
Molar mass of H2O = 18 g/mol
# moles of Fe = mass of Fe/molar mass Fe = 354/56 = 6.321 moles
# moles of H2O = mass of h2O/molar mass of H2O = 839/18 = 46.611 moles
Since moles of Fe is less than H2O; Fe is the limiting reagent.
Step 2: Calculate moles of Fe3O4 formed
As per reaction stoichiometry:
3 moles of Fe form 1 mole of Fe3O4
Therefore, 6.321 moles of Fe = 6.321 * 1/ 3 = 2.107 moles of Fe3O4
Step 4: calculate the mass of Fe3O4 formed
Molar mass of Fe3O4 = 232 g/mol
# moles = 2.107 moles
Mass of Fe3O4 = moles * molar mass
= 2.107 moles * 232 g/mol = 488.8 g (489 g approx)
