the reaction is
2NO(g) + 2H2(g) <—> N2(g) + 2H2O (g)
Kc = [N2] [ H2O]^2 / [NO]^2 [ H2]^2
Given
moles of NO = 0.124 therefore [NO] = moles /volume = 0.124 /2 = 0.062
moles of H2 = 0.0240 , therefore [H2] = moles / volume = 0.0240 / 2 = 0.012
moles of N2 = 0.0380 , therefore [N2] = moles / volume = 0.0380 / 2 = 0.019
moles of H2O = 0.0276 , therefore [H2O] = moles / volume = 0.0276 / 2 = 0.0138
Kc = (0.019) ( 0.0138)^2 / (0.062)^2 ( 0.012)^2 = 6.54
2.24 liters is the volume of the gas if pressure is increased to 1000 Torr.
Explanation:
Data given:
Initial volume of the gas V1 = 2.6 liters
Initial pressure of the gas P1 = 860 Torr 1.13 atm
final pressure on the gas P2 = 1000 Torr 1.315 atm
final volume of the gas after pressure change V2 =?
From the data given above, the law used is :
Boyles Law equation:
P1V1 = P2V2
V2 = P1V1/P2
= 1.13 X 2.6/ 1.31
= 2.24 Liters
If the pressure is increased to 1000 Torr or 1.315 atm the volume changes to 2.24 liters. Initially the volume was 2.6 litres and the pressure was 860 torr.
Answer:
The partial pressure of argon in the flask = 71.326 K pa
Explanation:
Volume off the flask = 0.001 
Mass of the gas = 1.15 gm = 0.00115 kg
Temperature = 25 ° c = 298 K
Gas constant for Argon R = 208.13 
From ideal gas equation P V = m RT
⇒ P = 
Put all the values in above formula we get
⇒ P = 
× 208.13 × 298
⇒ P = 71.326 K pa
Therefore, the partial pressure of argon in the flask = 71.326 K pa
I think its density is 850kg/m^3
Answer:
Sr would be the limiting reactant
5 moles
Explanation:
Since the equation is a balanced equation, the coefficient shows how each substance relates to the other in terms of the number of moles.
Reactants would be those on the left hand side of the arrow, while the products would be found on te right and side of the arrow. In this question, the reactants would be Sr and O₂.
Limiting reactant is the reactant that is insufficient; meaning to say that there is not enough of that substance and thus the reaction cannot continue. The other reactant(s) that is not limiting is called the excess reactants.
From the balanced equation, 2 moles of Sr is needed to react with 1 mole of O₂. Thus, if we have 5 moles of each reactant, Sr would be the limiting reactant since for every 1 mole of O₂, there has to be 2 moles of Sr in order for the reaction to proceed. Thus, if we have 5 moles of O₂, we would need 10 moles of Sr.
When we work out the amount of products formed, we look at the number of moles of the limiting reactant. This is because the limiting reactant determines how much is being reacted, while the excess number of moles of the excess reactant will remain unreacted.
For every 2 moles of Sr reacted, 2 moles of SrO would be produced. This means that the mole ratio of Sr to SrO is 1:1. Thus, since 5 moles of Sr has been reacted, 5 moles of the product (SrO) would be produced.
