The answer is 34.1 mL.
Solution:
Assuming ideal behavior of gases, we can use the universal gas law equation
P1V1/T1 = P2V2/T2
The terms with subscripts of one represent the given initial values while for terms with subscripts of two represent the standard states which is the final condition.
At STP, P2 is 760.0torr and T2 is 0°C or 273.15K. Substituting the values to the ideal gas expression, we can now calculate for the volume V2 of the gas at STP:
(800.0torr * 34.2mL) / 288.15K = (760.0torr * V2) / 273.15K
V2 = (800.0torr * 34.2mL * 273.15K) / (288.15K * 760.0torr)
V2 = 34.1 mL
The given equation from the problem above is already balance,
N2O5 ---> 2NO2 + 0.5O2
Since, in every mole of N2O5 consumed, 2 moles of NO2 are formed, we can answer the problem by multiplying the given rate, 7.81 mol/L.s with the ratio.
(7.81 mol/L.s) x (2 moles NO2 formed/ 1 mole of N2O5 consumed)
= 15.62 mol/L.s
The answer is the rate of formation of NO2 is approximately 15.62 mol/L.s.
One mole (abbreviated mol) is equal to 6.022×1023 molecular entities (Avogadro's number), and each element has a different molar mass depending on the weight of 6.022×1023 of its atoms (1 mole). The molar mass of any element can be determined by finding the atomic mass of the element on the periodic table.
There are a couple of ways todetermine if a reaction is exothermic or endothermic. Endothermic meaning that heat is added to the reaction to make the reactants interact and exothermic meaning heat is released during the reaction between the two reactants.
In endothermic reactions you can find a triangle above the arrow.
