The combustion reaction of octane is as follow,
C₈H₁₈ + 25/2 O₂ → 8 CO₂ + 9 H₂O
According to balance equation,
8 moles of CO₂ are released when = 114.23 g (1 mole) Octane is reacted
So,
6.20 moles of CO₂ will release when = X g of Octane is reacted
Solving for X,
X = (114.23 g × 6.20 mol) ÷ 8 mol
X = 88.52 g of Octane
Result:
88.52 g of Octane is needed to release 6.20 mol CO₂.
Answer:
6.05g
Explanation:
The reaction is given as;
Ethane + oxygen --> Carbon dioxide + water
2C2H6 + 7O2 --> 4CO2 + 6H2O
From the reaction above;
2 mol of ethane reacts with 7 mol of oxygen.
To proceed, we have to obtain the limiting reagent,
2,71g of ethane;
Number of moles = Mass / molar mass = 2.71 / 30 = 0.0903 mol
3.8g of oxygen;
Number of moles = Mass / molar mass = 3.8 / 16 = 0.2375 mol
If 0.0903 moles of ethane was used, it would require;
2 = 7
0.0903 = x
x = 0.31605 mol of oxygen needed
This means that oxygen is our limiting reagent.
From the reaction,
7 mol of oxygen yields 4 mol of carbon dioxide
0.2375 yields x?
7 = 4
0.2375 = x
x = 0.1357
Mass = Number of moles * Molar mass = 0.1357 * 44 = 6.05g
Answer:
The invention of the wheel had an important impact on farming.
The answer would be Linear. in Carbon Dioxide, the two double bonds count as two pairs of electrons around the carbon atom, predicting linear geometry.
source: http://chemistry.elmhurst.edu/vchembook/207epgeom.html
Answer:
171.34 g/mol
Explanation:
Ba molar mass = 137.328 g/mol
O molar mass = 15.999 g/mol * 2 = 31.9980 g/mol
H molar mass = 1.008 g/mol * 2 = 2.0160 g/mol
137.328 + 31.9980 + 2.0160 = 171.3420 = 171.34 g/mol
