Answer:
The answer is 6.25g.
Explanation:
First create your balanced equation. This will give you the stoich ratios needed to answer the question:
2C8H18 + 25O2 → 16CO2 + 18H2O
Remember, we need to work in terms of NUMBERS, but the question gives us MASS. Therefore the next step is to convert the mass of O2 into moles of O2 by dividing by the molar mass:
7.72 g / 16 g/mol = 0.482 mol
Now we can use the stoich ratio from the equation to determine how many moles of H2O are produced:
x mol H2O / 0.482 mol O2 = 18 H2O / 25 O2
x = 0.347 mol H2O
The question wants the mass of water, so convert moles back into mass by multiplying by the molar mass of water:
0.347 mol x 18 g/mol = 6.25g
Answer:

Explanation:
<em>Ferrous Sulphate</em>
<em> is generally found as Lime-Green Crystals. On heating, these crystals almost immediately turn white-yellow. They then, break down to produce an anhydrous mixture of Sulphur Trioxide </em>
<em>, Sulphur Dioxide </em>
<em> as well as Ferric Oxide </em>
<em>.</em>
<em>We can hence, frame a skeletal equation of this reaction and try to balance it.</em>
<em>Hence,</em>

<em>Now,</em>
<em>a)In order to balance it through the 'Hit &Trial Method', we'll follow a series of </em><em>steps</em><em>:</em>
<em>1. First, lets compare the number of Fe (Iron) atoms on the RHS and LHS. We find that, the no. of Fe Atoms on the RHS is twice the number of Fe Atoms on the LHS. We hence, add a co-effecient 2 beside </em>
.
<em>2. Now, Iron atoms, Sulphur Atoms and Oxygen atoms occur 2, 2, 8 respectively on both the sides:</em>
<em> Hence, As all the other elements as well as iron, balance, we've arrived upon our Balanced Equation :</em>
<em> </em>
<em>b) We know that, decomposition reactions are [generally] endothermic reactions in which Large Compounds </em><em>decompose </em><em>into smaller elements and compounds. Here, as Ferrous Sulphate </em><em>decomposes </em><em>into Sulphur Dioxide, Sulphur Trioxide and Ferric Oxide, the reaction that occurs here is </em><em>Decomposition Reaction.</em>
Answer:
1. NADH + H⁺ + FMN + Q ⟶ NAD⁺ + FMN + QH₂
2. The reactant that is reduced is Q
3. The charge on iron on the right side is +2, Fe²⁺
Explanation:
NADH + H⁺ + FMN + Q ⟶ NAD⁺ + FMN + QH₂
The reaction above is catalysed by NADH:ubiquinone oxidoreductase (complex 1), which transfers a hydride ion from NADH to FMN, from which two electrons pass through a series of of Fe-S centers to the iron-sulfur protein N-2. Electron transfer from N-2 to Ubiquinone forms QH₂
The species in a reaction which gains hydrogen irons is reduced, Therefore, the reactant that is reduced is Q, ubiquinone to form QH₂, ubiquinol.
To determine the oxidation number of iron on the right side of the reaction below,
QH2 + 2cyt c ( Fe3+) ⟶ Q + 2cyt c(Fex) + 2H^+
Sum of charges on the left side = Sum of charges on the right side
Sum of charges on the left side = 2 *+3 = +6
Therefore 2 * x + 2= 6
2x = 6 -2 = 4
x = 4/
x = 2
Therefore the charge on iron on the right side is +2, Fe²⁺
Answer:
About 0.652
Explanation:
Because the reaction is balanced, we can go straight to the next step. The molar mass of potassium is about 39.098, while the molar mass of hydrogen gas is 2 and the molar mass of water is 18. Therefore, 25.5g of potassium would be about 0.652 moles, and 220 grams of water would be about 12.222 moles, making potassium the limiting reactant. Since there is a single unit of each compound on both sides of the equation, there would be an equal amount of moles of potassium and hydrogen, and therefore about 0.652 moles of hydrogen gas would be produced. Hope this helps!