Answer:
They are both listed under group 11 on the periodic table and both are highly conductive of electricity
Explanation:
HOPE THIS HELPS ^^
Answer:
Mass = 42.8g
Explanation:
4 NH 3 ( g ) + 5 O 2 ( g ) ⟶ 4 NO ( g ) + 6 H 2 O ( g )
Observe that every 4 mole of ammonia requires 5 moles of oxygen to obtain 4 moles of Nitrogen oxide and 6 moles of water.
Step 1: Determine the balanced chemical equation for the chemical reaction.
The balanced chemical equation is already given.
Step 2: Convert all given information into moles (through the use of molar mass as a conversion factor).
Ammonia = 63.4g × 1mol / 17.031 g = 3.7226mol
Oxygen = 63.4g × 1mol / 32g = 1.9813mol
Step 3: Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio.
If all of the 1.9831 moles of oxygen were to be used up, there would need to be 1.9831 × 4 / 5 or 1.5865 moles of Ammonia. We have 3.72226 moles of ammonia - Far excess. Because there is an excess of Ammonia, the Oxygen amount is used to calculate the amount of the products in the reaction.
Step 4: Use the amount of limiting reactant to calculate the amount of H2O produced.
5 moles of O2 = 6 moles of H2O
1.9831 moles = x
x = (1.9831 * 6 ) / 5
x = 2.37972 moles
Mass of H2O = Molar mass * Molar mass
Mass = 2.7972 * 18
Mass = 42.8g
a. mass of iron = 69.92 g
b. percent yield = 93%
<h3>Further eplanation
</h3>
Percent yield is the compare of the amount of product obtained from a reaction with the amount you calculated
General formula:
Percent yield = (Actual yield / theoretical yield )x 100%
An actual yield is the amount of product actually produced by the reaction. A theoretical yield is the amount of product that you calculate from the reaction equation according to the product and reactant coefficients
a.
Reaction
Fe₂O₃+3CO⇒2Fe+3CO₂
MW Fe₂O₃ : 159.69 g/mol
mol Fe₂O₃
mol Fe₂O₃ : mol Fe = 1 : 2
mol Fe :
mass of Fe(Ar=55.845 g/mol) :
b.
actual yield = 65 g
theoretical yield = 69.92 g
percent yield :
