Answer:
102g
Explanation:
To find the mass of ethanol formed, we first need to ensure that we have a balanced chemical equation. A balanced chemical equation is where the number of atoms of each element is the same on both sides of the equation (reactants and products). This is useful as only when a chemical equation is balanced, we can understand the relationship of the amount (moles) of reactant and products, or to put it simply, their relationship with one another.
In this case, the given equation is already balanced.
From the equation, the amount of ethanol produced is twice the amount of yeast present, or the same amount of carbon dioxide produced. Do note that amount refers to the number of moles here.
Mole= Mass ÷Mr
Mass= Mole ×Mr
<u>Method 1: using the </u><u>mass of glucose</u>
Mr of glucose
= 6(12) +12(1) +6(16)
= 180
Moles of glucose reacted
= 200 ÷180
= 
mol
Amount of ethanol formed: moles of glucose reacted= 2: 1
Amount of ethanol
= 
= 
mol
Mass of ethanol
= ![\frac{20}{9} \times[2(12)+6+16]](https://tex.z-dn.net/?f=%5Cfrac%7B20%7D%7B9%7D%20%5Ctimes%5B2%2812%29%2B6%2B16%5D)
= 
= 102 g (3 s.f.)
<u>Method 2: using </u><u>mass of carbon dioxide</u><u> produced</u>
Mole of carbon dioxide produced
= 97.7 ÷[12 +2(16)]
= 97.7 ÷44
= 
mol
Moles of ethanol: moles of carbon dioxide= 1: 1
Moles of ethanol formed= mol
Mass of ethanol formed
= ![\frac{977}{440} \times[2(12)+6+16]](https://tex.z-dn.net/?f=%5Cfrac%7B977%7D%7B440%7D%20%5Ctimes%5B2%2812%29%2B6%2B16%5D)
= 102 g (3 s.f.)
Thus, 102 g of ethanol are formed.
Additional:
For a similar question on mass and mole ratio, do check out the following!
Answer:
Generally, the first ionisation energy increases along a period. But there are some exceptions one which is not an exception
10 x 70 = 100 x Part, or
700 = 100 x Part
Now, divide by 100 and get the answer:
Part = 700 / 100 = <span>7</span>
Answer:
Explanation:
The oxidation number is an integer that represents the number of electrons that an atom receives or makes available to others when it forms a given compound.
The oxidation number is positive if the atom loses electrons, or shares them with an atom that has a tendency to accept them. And it will be negative when the atom gains electrons, or shares them with an atom that has a tendency to give them up.
Chemical compounds are electrically neutral. That is, the charge that all the atoms of a compound contribute must be globally null. That is, when having positive or negative charges in a compound, their sum must be zero.
There are some rules for determining oxidation numbers in compounds. Among them it is possible to mention:
- Hydrogen (H) has an oxidation number +1 with nonmetals and - 1 with metals.
- Oxygen (O) presents the oxidation number -2
- Fluorine F has a unique oxidation state -1
Then:
- NOF: N+(-2)+(-1)=0 → N=3 → oxidation number of nitrogen (N) is +3, oxidation number of oxygen (O) is -2 and oxidation number of fluorine (F) is -1.
- ClF₅: Cl + 5*(-1)=0 → Cl= 5 → oxidation number of chlorine (Cl) is +5 and oxidation number of fluorine (F) is -1.
- H₂SO₃: 2*(+1)+S+3*(-2)=0 → S=4 → oxidation number of hydrogen (H) is +1, oxidation number of oxygen (O) is -2 and oxidation number of sulfur (S) is +4.
Answer:
1. C4H8 + 6O2 -----> 4CO2 + 4H20
2. 3836.77 kcal
Explanation:
1. Balanced equation for the complete combustion of cyclobutane:
C4H8 + 6O2 -----> 4CO2 + 4H20
2. Heat of combustion of cyclobutane = 650.3 kcal/mol
Molecular weight of cyclobutane, C4H8 = 56.1 g/mol
Mole of C4H8 : mass of cyclobutane/Molecular weight of cyclobutane
Mole of C4H8 = 331/56.1 = 5.9 mol
Energy released during combustion = 5.9 mol × 650.3 kcal/mol = 3836.77kcal
Therefore the energythat is released during the complete combustion of 331 grams of cyclobutane is 3836.77kcal
