Answer:
A) 8.00 mol NH₃
B) 137 g NH₃
C) 2.30 g H₂
D) 1.53 x 10²⁰ molecules NH₃
Explanation:
Let us consider the balanced equation:
N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)
Part A
3 moles of H₂ form 2 moles of NH₃. So, for 12.0 moles of H₂:
Part B:
1 mole of N₂ forms 2 moles of NH₃. And each mole of NH₃ has a mass of 17.0 g (molar mass). So, for 4.04 moles of N₂:
Part C:
According to the <em>balanced equation</em> 6.00 g of H₂ form 34.0 g of NH₃. So, for 13.02g of NH₃:
Part D:
6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂:
Answer:
its very simple ans we have 2 just multiply256
Get to know first how many moles in the gas:n = pV/RT= (1.013*10^5*750/760) Pa *1.49*10^-3 m^3/(8.314 J/(molK)*298) n = 0.0601 moles.
The combustion energies are 889 kJ/mol (methane) and 2 220 kJ (propane) x = moles methane, y = moles propane
x*889 + y*2220 = 778 x + y = 0.0601----------- x = 0.267784 moles = 0.267784*100/0.0601 = 44.6 % y = 0.243216 moles = 0.243216*100/0.0601 = 55.4 %
Answer:
6.64
Explanation:
so moles
then do =4.01×10^24/6.02×10^23
you get the answer as 6.6445 u only need to write the answer to
3.S.F 6.64 moles
hope this make sense :)