Answer:
D.
Explanation:
That would be the last option.
Repeated similar experimental outcomes are confirmation of the theory.
Answer:
Explanation:
Mole = no. Molecules/6.02×10^23
Mole = (2.4×10^25)/(6.02×10^23)
Mole = 39.87mole
Molar mass of NaHCO3 is
= 23 + 1 + 12 + 3(16)
= 84g/mol
Mole = mass/molar mass
Mass = Mole × molar mass
Mass = 39.87 × 84
Mass= 3348.84g
- Iron is in limited amount, limiting reagent
- The theoretical yield of iron(III) oxide formed is 27.1 grams.
Given:
The reaction between iron and dioxygen at high temperatures to form iron(III) oxide.
To find:
- The limiting reagent
- The theoretical yield of iron(III) oxide formed when Suppose 19.1 g
of iron is reacted with 17.9 g of oxygen.
Solution:
- The mass of iron = 19.1 g
- The moles of iron
- The mass of dioxygen = 19.1 g
- The moles of dioxygen
According to reaction, 4 moles of iron reacts with 3 moles of dioxygen then 0.342 moles of iron will react with:
0.2465 moles of dioxygen will react with 0.342 moles of iron which means that:
- Iron is in limited amount, limiting reagent
- dioxygen is in an excessive amount, excessive reagent.
- The amount of iron(III) oxide formed will depend upon moles of iron.
According to reaction, 4 moles of iron gives 2 moles of iron (III) oxide, then 0.342 moles of iron will give:
The theoretical yield of 0.171 moles of iron(III) oxide:
The theoretical yield of iron(III) oxide formed is 27.1 grams.
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