The balanced equation for the reaction is as follows
2H₂ + O₂ --> 2H₂O
stoichiometry of H₂ to O₂ is 2:1
number of H₂ moles - 30.0 g / 2 g/mol = 15 mol
number of O₂ moles - 80.0 g / 32 g/mol = 2.5 mol
limiting reactant is the reagent in which only a fraction is used up in the reaction
if H₂ is the limiting reactant
if 2 mol of H₂ requires 1 mol of O₂
then 15 mol of H₂ requires 1/2 x 15.0 = 7.5 mol of O₂
but only 2.5 mol of O₂ is required
this means that O₂ is the limiting reagentt and H₂ is in excess
Answer:-
2328.454 grams
Explanation:-
Volume V = 18.4 litres
Temperature T = 15 C + 273 = 288 K
Pressure P = 1.5 x 10^ 3 KPa
We know universal Gas constant R = 8.314 L KPa K-1 mol-1
Using the relation PV = nRT
Number of moles of oxygen gas n = PV / RT
Plugging in the values
n = (1.5 x 10^3 KPa ) x ( 18.4 litres ) / ( 8.314 L KPa K-1 mol-1 x 288 K)
n = 11.527 mol
Now the balanced chemical equation for this reaction is
2KNO3 --> 2KNO2 + O2
From the equation we can see that
1 mol of O2 is produced from 2 mol of KNO3.
∴ 11.527 mol of O2 is produced from 2 x 11.527 mol of KNO3.
= 23.054 mol of KNO3
Molar mass of KNO3 = 39 x 1 + 14 x 1 + 16 x 3 = 101 grams / mol
Mass of KNO3 = 23.054 mol x 101 gram / mol
= 2328.454 grams
Electron - negligible mass, negative charge, orbits the nucleus
Proton - 1 AMU, positive charge, in the nucleus
Neutron, 1 AMU, no charge, in the nucleus
Answer: I think the answer is 2.5 M.
Explanation: Formula: M = n/v
M= 10 mol / 4L
M= 2.5 M
