mass of pentane : = 30.303 g
moles of Al₂(CO₃)₃ : = 0.147
<h3>Further explanation</h3>
Given
1. Reaction
C₅H₁₂+8O₂→6H₂O+5CO₂.
45.3 g water
2. 2AlCl₃ + 3MgCO₃ → Al₂(CO₃)₃ + 3MgCl₂
37.2 MgCO₃
Required
mass of pentane
moles of Al₂(CO₃)₃
Solution
1. mol water = 45.3 : 18 g/mol = 2.52
From equation, mol ratio of C₅H₁₂ : H₂O = 1 : 6, so mol pentane :
= 1/6 x mol H₂O
= 1/6 x 2.52
= 0.42
Mass pentane :
= mol x MW
= 0.42 x 72.15 g/mol
= 30.303 g
2. mol MgCO₃ : 37.2 : 84,3139 g/mol = 0.44
mol Al₂(CO₃)₃ :
= 1/3 x mol MgCO₃
= 1/3 x 0.44
= 0.147
Answer:
54 g is the theoretical yield
Explanation:
This is the reaction:
2H₂ + O₂ → 2H₂O
So 2 moles of hydrogen react with 1 mol of oxygen, to produce 2 mol of water.
If I have 3 moles of H₂ and 2 moles of O₂, the my limiting reactant is the hydrogen.
1 mol of O₂ react with 2 moles of H₂
S 2 mol of O₂ would react with 4 moles (I only have 3 moles)
Then, ratio is 2:2 the same as 1:1
As 2 mol of H₂ produce 2 moles of water, 3 moles of H₂ will produce 3 moles of H₂O.
This is the theoretical yield in moles. Let's convert them to mass (mol . molar mass)
3 mol . 18g/m = 54 g
From what i can gather it looks like d
Answer:
Explanation:
The combustion reaction of Octane is:
To calculate the mass of CO₂ and H₂O produced, we need to know the mass of octane combusted.
We calculate the mass of Octane from the given volume and density, using the following <em>conversion factors</em>:
Now we<u> convert 1.24 gallons to mL</u>:
- 1.24 gallon *
4693.4 mL
We <u>calculate the mass of Octane</u>:
- 4693.4 mL * 0.703 g/mL = 3.30 g Octane
Now we use the <em>stoichiometric ratios</em> and <em>molecular weights</em> to <u>calculate the mass of CO₂ and H₂O</u>:
- CO₂ ⇒ 3.30 g Octane ÷ 114g/mol *
* 44 g/mol = 10.19 g CO₂
- H₂O ⇒ 3.30 g Octane ÷ 114g/mol *
* 18 g/mol = 4.69 g H₂O
Explanation:
the proper format for scientific notation the following conditions must be true
