Answer:
0.703g Na must reacted
Explanation:
The reaction of Sodium, Na, With water, H₂O is:
2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g)
<em>Where 2 moles of sodium reacts with an excess of water to produce 1 mole of hydrogen</em>
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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:
<em>Moles H₂:</em>
PV = nRT
PV /RT = n
<em>Where P is pressure = 1.15atm</em>
<em>V is volume in liters = 0.325L</em>
<em>R is gas constant = 0.082atmL/molK</em>
<em>T is absolute temperature = 25°C + 273 = 298.0K</em>
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1.15atm*0.325L / 0.082atmL/molK*298.0K = n
0.0153 moles of hydrogen are produced
<em>Moles Na:</em>
0.0153 moles H₂ * (2moles Na / 1mol H₂) = 0.0306 moles Na
<em>Mass Na -Molar mass: 22.99g/mol-:</em>
0.0306 moles Na * (22.99g / mol) =
<h3>0.703g Na must reacted</h3>
