Electromagnetic radiation emitted from a hot object reaches it's peak at a wavelength of 8.5 x 10-7 m. What is the frequency of
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The question is incomplete. The complete question is :
Hydrogen gas and oxygen
gas react to form water vapor
. Suppose you have 11.0 mol of
and 13.0 mol of
in a reactor. Calculate the largest amount of
that could be produced. Round your answer to the nearest 0.1 mol .
Solution :
The balanced reaction for reaction is :
11.0 13.0
11/2 13/1 (dividing by the co-efficient)
6.5 mol 13 mol (minimum is limiting reagent as it is completely consumed during the reaction)
Therefore, is limiting reagent. It's stoichiometry decides the product formation amount from equation above it is clear that number of moles for
will be produced = number of moles of
= 11.0 mol
Answer:
0.9401 mol H₂O
General Formulas and Concepts:
<u>Chemistry - Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
16.94 g H₂O
<u>Step 2: Identify Conversions</u>
Molar Mass of H - 1.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol
<u>Step 3: Convert</u>
<u /> = 0.940067 mol H₂O
<u>Step 4: Check</u>
<em>We are given 4 sig figs. Follow sig fig rules and round.</em>
0.940067 mol H₂O ≈ 0.9401 mol H₂O
<u>Answer:</u> The moles of silver nitrate reacted is 5.72 moles, moles of iron (II) nitrate produced is 2.86 moles and moles of silver chloride produced is 5.72 moles
<u>Explanation:</u>
We are given:
Moles of iron (II) chloride = 2.86 moles
For the given chemical equation:
- <u>For silver nitrate:</u>
By Stoichiometry of the reaction:
1 mole of iron (II) chloride reacts with 2 moles of silver nitrate.
So, 2.86 moles of iron (II) chloride will react with = of silver nitrate
Moles of silver nitrate reacted = 5.72 moles
- <u>For iron (II) nitrate:</u>
By Stoichiometry of the reaction:
1 mole of iron (II) chloride produces 1 mole of iron (II) nitrate
So, 2.86 moles of iron (II) chloride will produce = of iron (II) nitrate
Moles of iron (II) nitrate produced = 2.86 moles
- <u>For silver chloride:</u>
By Stoichiometry of the reaction:
1 mole of iron (II) chloride produces 2 moles of silver chloride
So, 2.86 moles of iron (II) chloride will produce = of silver chloride
Moles of silver chloride produced = 5.72 moles
Hence, the moles of silver nitrate reacted is 5.72 moles, moles of iron (II) nitrate produced is 2.86 moles and moles of silver chloride produced is 5.72 moles