Mass is density times volume
0.760 x 531 = 403.56g
404g if 3 sig figs
Answer:
V = 22.34 L
Explanation:
Given data:
Volume of O₂ needed = ?
Temperature and pressure = standard
Number of molecules of water produced = 6.0× 10²³
Solution:
Chemical equation:
2H₂ + O₂ → 2H₂O
Number of moles of water:
1 mole contain 6.022× 10²³ molecules
6.0× 10²³ molecules × 1 mole / 6.022× 10²³ molecules
0.99 mole
Now we will compare the moles of oxygen and water.
H₂O : O₂
2 : 1
0.996 : 0.996
Volume of oxygen needed:
PV = nRT
V = nRT/P
V = 0.996 mol × 0.0821 atm.L/mol.K × 273.15 K / 1 atm
V = 22.34 L
