Answer:
11.3 g of H₂O will be produced.
Explanation:
The combustion is:
2C₈H₁₈ + 25O₂→ 16CO₂ + 18H₂O
First of all, we determine the moles of the reactants in order to find out the limiting reactant.
8 g / 114g/mol = 0.0701 moles of octane
37g / 32 g/mol = 1.15 moles of oxygen
The limiting reagent is the octane. Let's see it by this rule of three:
25 moles of oxygen react to 2 moles of octane so
1.15 moles of oxygen will react to ( 1.15 . 2)/ 25 = 0.092 moles of octane.
We do not have enough octane, we need 0.092 moles and we have 0.0701 moles. Now we work with the stoichiometry of the reaction so we make this rule of three:
2 moles of octane produce 18 moles of water
Then 0.0701 moles of octane may produce (0.0701 . 18)/2= 0.631 moles of water.
We convert the moles to mass → 0.631 mol . 18 g/1mol = 11.3 g of H₂O will be produced.
Answer:
B and C is your answer
Explanation:
Hope I helped, Sorry if I'm wrong
Answer:
The answer is D. 0.60 L
Explanation:
The balanced reaction equation including states of matter is;
H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)
More simple:
H2SO4 + 2NaOH → Na2SO4 + 2H2O
Now, we can see from this reaction equation that the mole ratio of NaOH to H2SO4 is 2:1
Number of moles of H2SO4 reacted = 1.2 moles
Hence;
2 moles of NaOH reacts with 1 mole of H2SO4
x moles of NaOH reacts with 1.2 moles of H2SO4
x = 2 * 1.2/1 = 2.4 moles of NaOH
Recall that;
Number of moles = Concentration * Volume
Volume = number of moles/concentration
Volume of NaOH is obtained from;
Volume = 2.4 moles/ 4.0 M
Volume = 0.60 L
Answer:
20cm^2
Explanation:
Here, Density= Mass/ Volume
=100/5
= 20 cm^2
Answer:
20 g Ag
General Formulas and Concepts:
<u>Chemistry - Stoichiometry</u>
- Using Dimensional Analysis
<u>Chemistry - Atomic Structure</u>
Explanation:
<u>Step 1: Define</u>
[RxN] Cu (s) + AgNO₃ (aq) → CuNO₃ (aq) + Ag (s)
[Given] 10 g Cu
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol Cu = 1 mol Ag
Molar Mass of Cu - 63.55 g/mol
Molar Mass of Ag - 197.87 g/mol
<u>Step 3: Stoichiometry</u>
<u />
= 16.974 g Ag
<u>Step 4: Check</u>
<em>We are given 1 sig fig. Follow sig fig rules and round.</em>
16.974 g Ag ≈ 20 g Ag
