Hey there!:
Molar mass N₂ = 28.0134 g/mol
28.0134 g ------------------- 22.4 L (at STP )
mass N₂ -------------------- 50.0 L
mass N₂ = 50.0 x 28.0134 / 22.4
mass N₂ = 1400.67 / 22.4
mass N₂ = 62.529 g
Hope this helps!
Answer:
7?
Explanation:
Its somewhat hard to comprehend the question, but if the way I read it was right, its 7.
Answer:
The answer to your question is: 65.9 g released of CO2
Explanation:
MW CO2 = 44 g
MW CuCO3 = 123.5 g
CO2 released = ?
CuCO3 = 185 g
CuCO3 ⇒ CO2 + CuO
123.5 ----------- 44g
185 g ----------- x
x = (185 x 44) / 123.5
x = 65.9 g released of CO2
4.06x20^24/6.02x10^23 = 6.744 moles x 55.845 g/mole = 376.61868grams
Answer:
2Al+1.5O2→Al2O3
Thus, 2 mol of Al combine with 1.5 mol of oxygen to form 1 mol of Al2O3.
2 mol of Al corresponds to 2×27=54 g.
Thus, the weight of Al used in the reaction is 54 g.