Answer:
0.703g Na must reacted
Explanation:
The reaction of Sodium, Na, With water, H₂O is:
2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g)
Where 2 moles of sodium reacts with an excess of water to produce 1 mole of hydrogen
To solve this question we have to use PV = nTR to solve the moles of the gas. With the moles of hydrogen we can find the moles of sodium that reacted and its mass:
Moles H₂:
PV = nRT
PV /RT = n
Where P is pressure = 1.15atm
V is volume in liters = 0.325L
R is gas constant = 0.082atmL/molK
T is absolute temperature = 25°C + 273 = 298.0K
1.15atm*0.325L / 0.082atmL/molK*298.0K = n
0.0153 moles of hydrogen are produced
Moles Na:
0.0153 moles H₂ * (2moles Na / 1mol H₂) = 0.0306 moles Na
Mass Na -Molar mass: 22.99g/mol-:
0.0306 moles Na * (22.99g / mol) =
