A chemist combines 11.3 L N2 gas, 34.3 L O2 gas, and 17.5 L Br2 gas in a reaction vessel at STP. The gases react according to the following equation.
N2(g) + O2(g) + Br2(g) → 2 NOBr(g)
How many moles of each gas are initially present in the vessel, before any reaction occurs?
N2 ___ | mol |
O2 ___ | mol |
Br2 ___ | mol |
Compound | Molar Mass g/mol | Density g/L at 298 K |
N2 | 28.014 | 1.25 |
O2 | 32 | 1.429 |
Br2 | 159.81 | 3100 |
VN2 = 11.3 L
So, Mass of N2 = VN2 * density = 11.3 L * 1.25 g/L =14.125 g
Moles of N2 = Mass /Molar mass= 14.125/28.014 = 0.504 mol
So, Moles of N2 = 0.504 mol
VO2= 34.3 L
So, Mass of O2 = VO2 * density = 34.3 L * 1.429 g/L =49.015 g
Moles of O2 = Mass /Molar mass= 49.0155/32 = 1.532 mol
So, Moles of O2 = 1.532 mol
VBr2 = 17.5 L
So, Mass of Br2 = VBr2 * density = 17.5 L * 3100 g/L =54250 g
Moles of Br2 = Mass /Molar mass= 54250/159.81 = 339.47 mol
So, Moles of Br2 = 339.47 mol
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