A piece of sodium metal reacts completely with water as follows: 2Na(s)+2H2O(l)⇒2NaOH(aq)+H2(g) The hydrogen gas generated is collected over water at 25.0 degree C. The volume of the gas is 246 mL measured at 1.00 atm. Calculate the number of grams of sodium used in the reaction. (Vapor pressure of water at 25 degree C=0.0313 atm.)

Question

A piece of sodium metal reacts completely with water as follows:   2Na(s)+2H2O(l)⇒2NaOH(aq)+H2(g)   The hydrogen gas generated is collected over water at 25.0 degree C. The volume of the gas is 246 mL measured at 1.00 atm. Calculate the number of grams of sodium used in the reaction. (Vapor pressure of water at 25 degree C=0.0313 atm.)

Daniel Cormack · Accepted Answer

Step 1 Given information: The equation for the reaction is 2Na(s)+2H2O(l)⇒2NaOH(aq)+H2(g) −T=25.0∘C=(273+25)K=298K; −V=246mL=0.246L is the volume of the gas (H2); −P−1.00atm is the total pressure over the water; - R=0.0821LatmKmol is the ideal gas constant; - PH2O=0.0313atm at 25.0∘C is the vapor pressure of water. Let&#039;s calculate the number of grams of sodium used in the reaction. Step 2 The total pressure of the mixture is the summary of partial pressures of each component. P{2}+PH2O=PTotal First, let&#039;s find the partial pressure of the H2: P{2}=PTotal−PH2O PH2=1.00atm−0.0313atm PH2=0.9687atm Now let&#039;s calculate the number of moles of H2. We can use the ideal gas equation. PV=nRT n=PVRT n=0.9687atm⋅0.246L0.0821L atmK mol⋅298K n=9.740⋅10−3mol The number of moles of H2 is9.740⋅10−3mol. Step 3 From the balanced equation for the chemical reaction above, we can conclude that 2mol of Na produces 1mol of H2. Therefore, number of moles of Na which produce H2 is 9.740⋅10−3mol is: nNa=2molNa1molH2⋅9.740⋅10−3molH2 nNa=0.01948molNa Now we can find the mass of sodium in g used in this reaction. nNa=mNaMNa Therefore, mNa=nNa⋅M

twineg4 · Accepted Answer

Explanation:&amp;nbsp;1. Collect all of the information in one location.Mr:22.99&amp;nbsp;2Na+2H2O⇒2NaOH+H2&amp;nbsp;P→t=1.00atm&amp;nbsp;pH2O=0.0313atm&amp;nbsp;T=25.0∘C&amp;nbsp;V=246mL&amp;nbsp;2. Moles of&amp;nbsp;H2&amp;nbsp;We can use the Ideal Gas Law to calculate the moles of hydrogen:pV=nRT&amp;nbsp;(a) Determine the hydrogen partial pressure.&amp;nbsp;p→t=pH2+pH2O&amp;nbsp;1.00atm=pH2+0.0313atm&amp;nbsp;pH2=0.9687atm&amp;nbsp;(b) Convert the volume to litres&amp;nbsp;V=246mL=0.246L&amp;nbsp;(c) Convert the temperature to kelvins&amp;nbsp;T=(25.0+273.15)K=298.15K&amp;nbsp;(d) Calculate the moles of hydrogen&amp;nbsp;0.9687atm×0.246L=n×0.08206L⋅atm⋅K−1mol−1×298.15K&amp;nbsp;0.2383=24.47nmol−1&amp;nbsp;n=0.238324.47mol−1&amp;nbsp;=0.009740mol&amp;nbsp;3. Moles of Na&amp;nbsp;The molar ratio is 2 mol Na: 1 mol&amp;nbsp;H2&amp;nbsp;Moles of&amp;nbsp;Na=0.009&amp;nbsp;740molH2×2molNa1molH2=0.01948molNa&amp;nbsp;4, Mass of Na&amp;nbsp;Mass of&amp;nbsp;Na=0.019&amp;nbsp;48molNa×22.99gNa1molNa=0.449gNa&amp;nbsp;The Na mass used was 0.449g.

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