Answer:
21.6 g
Explanation:
The reaction that takes place is:
First we<u> convert the given masses of both reactants into moles</u>, using their <em>respective molar masses</em>:
- 9.6 g CH₄ ÷ 16 g/mol = 0.6 mol CH₄
- 64.9 g O₂ ÷ 32 g/mol = 2.03 mol O₂
0.6 moles of CH₄ would react completely with (2 * 0.6) 1.2 moles of O₂. As there are more O₂ moles than required, O₂ is the reactant in excess and CH₄ is the limiting reactant.
Now we <u>calculate how many moles of water are produced</u>, using the <em>number of moles of the limiting reactant</em>:
- 0.6 mol CH₄ *
= 1.2 mol H₂O
Finally we<u> convert 1.2 moles of water into grams</u>, using its <em>molar mass</em>:
- 1.2 mol * 18 g/mol = 21.6 g
Answer:
A
Explanation:
bc inorganic compoud refers to all compound that do not contain carbons.
Answer:
So the volume will be 2.33 L
Explanation:
The reaction for the combustion is:
2 C₄H₁₀ (g) + 13 O₂ (g) → 8 CO₂ (g) + 10 H₂O (l)
mass of butane to moles (mass / molar mass)
1.4 g / 58 g/mol
= 0.024 moles
2 moles of butane can produce 8 moles of carbon dioxide
0.024 moles of butane must produce (0.024 × 8) /2
= 0.096 moles of CO₂
Now we apply the Ideal Gases Law to find out the volume formed.
P . V = n . R . T
p = 1atm
n = 0.096 mol
R = 0.082 L.atm/mol.K
T = 273 + 23 = 296K
V = ?
1atm × V = 0.096 mol × 0.082 L.atm/mol.K × 296K
V = 0.096 mol × 0.082 L.atm/mol.K × 296K / 1atm
= 2.33 L
So the volume will be 2.33 L
Answer:
A
Explanation:
CH3CH2COOH- Propanoic Acid