Answer:
54.4 mol
Explanation:
the equation for complete combustion of butane is
2C₄H₁₀ + 13O₂ ---> 8CO₂ + 10H₂O
molar ratio of butane to CO₂ is 2:8
this means that for every 2 mol of butane that reacts with excess oxygen, 8 mol of CO₂ is produced
when 2 mol of C₄H₁₀ reacts - 8 mol of CO₂ is produced
therefore when 13.6 mol of C₄H₁₀ reacts - 8/2 x 13.6 mol = 54.4 mol of CO₂ is produced
therefore 54.4 mol of CO₂ is produced
I'm assuming that C is carbon.
55.1259 g of C
I’m pretty sure it’s A or C
Answer:
17.5moles
Explanation:
The number of moles in a substance can be calculated by using the formula;
Number of moles (n) = mass (m) ÷ molar mass (MM)
According to this question, mass of ammonia (NH3) = 297g
Molar Mass of NH3 = 14 + 1(3)
= 17g/mol
n = 297/17
n = 17.47
Number of moles of NH3 = 17.5moles
Answer:
V = 27.98 L
Explanation:
Given data:
Mass of CO₂ = 33.0 g
Pressure = 500 torr
Temperature = 27°C
Volume occupied = ?
Solution:
Number of moles of CO₂:
Number of moles = mass/molar mass
Number of moles = 33.0 g/ 44 g/mol
Number of moles = 0.75 mol
Volume of CO₂:
PV = nRT
R = general gas constant = 0.0821 atm.L/ mol.K
Now we will convert the temperature.
27+273 = 300 K
Pressure = 500 /760 = 0.66 atm
By putting values,
0.66 atm×V = 0.75 mol × 0.0821 atm.L/ mol.K × 300 K
V = 18.47 atm.L/0.66 atm
V = 27.98 L
