Answer:
70mol
Explanation:
The equation of the reaction is given as:
2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O
Given parameters:
Number of moles of acetylene = 35.0mol
Number of moles of oxygen in the tank = 84.0mol
Unknown:
Number of moles of CO₂ produced = 35.0mol
Solution:
From the information given about the reaction, we know that the reactant that limits this combustion process is acetylene. Oxygen is given in excess and we don't know the number of moles of this gas that was used up. We know for sure that all the moles of acetylene provided was used to furnish the burning procedure.
To determine the number of moles of CO₂ produced, we use the stoichiometric relationship between the known acetylene and the CO₂ produced from the balanced chemical equation:
From the equation:
2 moles of acetylene produced 4 moles of CO₂
∴ 35.0 mol of acetylene would produced:
= 70mol
Cu + S ---> CuS
by reaction 1 mol 1 mol
from the problem 0.25 mol 0.25 mol
0.25 mol Cu
Answer:
im pretty sure its b
Explanation:
im sorry if im wrong i tried my best
Answer:
Explanation:
C = 41.4/12 = 3.43
H = 3.47/1 = 3.47
O = 55.1/16 =3.44
CHO is the skeletal formula (divide each by the lowest number above). The results are close enough to 1 to be 1.
(CHO)_x = 116
C + H + O = 29
(29) _ x = 116
x = 116/29
x = 4
So there area 4 carbons 4 hydrogens and 4 oxygens.
The correct formula is C4H4O4
Answer:
40.94 g
Explanation:
Given data:
Mass of NO₂ = ?
Volume = 20.0 L
Pressure = 110.0 Pka
Temperature = 25°C
Solution:
Pressure = 110.0 KPa (110/101 = 1.1 atm)
Temperature = 25°C (25+273 = 298.15 K)
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
n = PV/RT
n = 1.1 atm × 20.0 L / 0.0821 atm.L/ mol.K ×298.15 K
n = 22 / 24.5 /mol
n= 0.89 mol
Mass of NO₂:
Mass = number of moles × molar mass
Mass = 0.89 mol × 46 g/mol
Mass = 40.94 g
