Answer:
HNO3 + KOH → _KNO3___ + __H2O___
Explanation:
It's balanced
hope it helps
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:
11 molecules of CH4.
23 atoms of C is the leftover.
Explanation:
Hello!
In this case, for the formation of methane:
We can see there is an excess of carbon based on their stoichiometry, because the needed amount of hydrogen gas molecules would be:
Thus, the formed molecules of methane are computed below:
In such a way, the leftover of carbon atoms are:
Best regards!
Answer:
The answer to your question is 0.4 moles of Oxygen
Explanation:
Data
Octane (C₈H₈)
Oxygen (O₂)
Carbon dioxide (CO₂)
Water (H₂O)
moles of water = ?
moles of Oxygen = 1
Balanced chemical reaction
C₈H₈ +10O₂ ⇒ 8CO₂ + 4H₂O
Reactant Element Products
8 C 8
8 H 8
20 O 20
Use proportions to solve this problem
10 moles of Oxygen ----------------- 4 moles of water
1 mol of Oxygen ------------------ x
x = (4 x 1) / 10
x = 4 / 10
x = 0.4 moles of water
