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:
B I have taken the quiz already 90%
To solve this problem we just need to use the rule of three:
150g..................395.1J
450g................xJ
x = 450*395.1/150 = 1185,3J
450.0 g of the substance completely reacted with oxygen will produce 1.1853 kJ(<span>kiloJoule</span>)
From the given balanced equation we have find out the amount (in gm) of Ag formed from 5.50 gm of Ag₂O.
2Ag₂O(s) → 4Ag (s) + O₂ (g)
We know, molecular mass of Ag₂O= 231.7 g/mol, and atomic mass of Ag= 107.8 g/mol. Given, mass of Ag₂O=5.50 gm. Number moles of Ag₂O=
= 0.0237 moles.
From the balanced chemical reaction we get 2 (two) moles of Ag₂O produces 4 (four) moles of Ag. So, 0.0237 moles of Ag₂O produces 
moles=0.0474 moles of Ag= 0.0474 X 107.8 g of Ag=5.11g Ag.
Therefore, 5.50 g Ag₂O produces 5.11 g of Ag as per the given balanced chemical reaction.
