Determine the mass of water formed when 12.5 L NH3 (at 298 K and 1.50 atm) is reacted with 18.9 L of O2 (at 323 K and 1.1 atm). 4NH3(g)+5O2(g)⇒4NO(g)6H2O(g)

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Determine the mass of water formed when 12.5 L NH3 (at 298 K and 1.50 atm) is reacted with 18.9 L of O2 (at 323 K and 1.1 atm).   4NH3(g)+5O2(g)⇒4NO(g)6H2O(g)

Cheryl King · Accepted Answer

Step 1  The balanced reaction taking place is given as,  ⇒4NH3(g)+5O2(g)⇒4NO(g)+6H2O(g)  According to ideal gas law,  PV=nRT  where P= pressure in atm   V= volume in L  n= moles  R=gas constant=0.0821 atm.LK.mol  T=temperature in K=T∈C∘+273  Step 2  a) For NH3,P=1.50 atm,V=12.5L and T=298K  Hence substituting the values we get,  ⇒1.50×12.5=n×0.0821×298   ⇒n=moles of NH3 taken=0.7664 mol approx.  b) For O2,P=1.10 atm,V=18.9LandT=323K  Hence substituting the values we get,  ⇒1.10×18.9=n×0.0821×323  ⇒n=moles of O2 taken=0.784 mol approx.  Step 3  ⇒4 NH3(g)+5O2(g)⇒4 NO(g)+6H2O(g)  From the above balanced reaction we can see that 5 moles of O2 is required to react with 4 moles of NH3   Hence moles of O2 required =moles of NH3 taken×54=0.7664×54=0.958 mol  Since moles of O2 taken is less than required. Hence O2 is the limiting reagent and will react completely.  From the above reaction we can see that 5 moles of O2 is reacting to produce 6 moles of H2O.  Hence moles of H2O that can be produced =moles of O2 taken×65=0.784×65=0.9408 mol  Molar mass of H2O=Atomic mass of H×2+Atomic mass of O=1×2+16=18gmol  Hence mass of H2O that can be produced =moles×molar mass=0.9804×18=16.9g approx.  Hence the answer is 16.9 g.

Mary Herrera · Accepted Answer

The mass of water formed is calculation Use the ideal gas equation to calculate the moles of NH3 and O2 that is Pv=nRT where; P= pressure, V= volume, n= number of moles, R=gas constant=0.0821l.atmmol.K make n the formula of the subject by diving both side by RT n=PVRT The moles of NH3 n=1.50atm×12.5L0.0821L.atmmol.k×268K=0.766 moles The moles of O2 =1.1atm×18.9L0.0821L.atmmol.k×323K=0.784 moles write the reaction between NH3 and O2 4NH3+5O2⇒4 No+6H2O from equation above 0.766 moles of NH3 reacted to produce 0.766×64=1.149 moles of H2O 0.784moles of O2 reacted to produce0.784×65=0.9408moles of H20 since O2 is totally consumed, O2 is the limiting reagent and therefore the moles of H2O produced =0.9408 moles mass of H2O=moles×molar mass from periodic table the molar mass of H2O=(1×2)+16=18gmol mass=18gmol×0.9408moles=16.93grams

nick1337 · Accepted Answer

Solution:First, let&#039;s convert the given volumes of NH3 and O2 to the number of moles using the ideal gas law:For NH3:PV=nRTwhere:P = pressure of NH3 = 1.50 atmV = volume of NH3 = 12.5 Ln = number of moles of NH3R = ideal gas constant = 0.0821 L atm/(mol K)T = temperature of NH3 = 298 KRearranging the equation to solve for n:n=PVRTSubstituting the given values:nNH3=(1.50&amp;nbsp;atm)(12.5&amp;nbsp;L)(0.0821&amp;nbsp;L&amp;nbsp;atm/(mol&amp;nbsp;K))(298&amp;nbsp;K)Now, let&#039;s calculate the number of moles of O2 using the same approach:For O2:PV=nRTwhere:P = pressure of O2 = 1.1 atmV = volume of O2 = 18.9 Ln = number of moles of O2R = ideal gas constant = 0.0821 L atm/(mol K)T = temperature of O2 = 323 KRearranging the equation to solve for n:n=PVRTSubstituting the given values:nO2=(1.1&amp;nbsp;atm)(18.9&amp;nbsp;L)(0.0821&amp;nbsp;L&amp;nbsp;atm/(mol&amp;nbsp;K))(323&amp;nbsp;K)Next, let&#039;s determine the limiting reactant by comparing the number of moles of NH3 and O2. The balanced equation shows that 4 moles of NH3 react with 5 moles of O2. Therefore, the ratio of moles is 4:5.We have:nNH3=(1.50&amp;nbsp;atm)(12.5&amp;nbsp;L)(0.0821&amp;nbsp;L&amp;nbsp;atm/(mol&amp;nbsp;K))(298&amp;nbsp;K)nO2=(1.1&amp;nbsp;atm)(18.9&amp;nbsp;L)(0.0821&amp;nbsp;L&amp;nbsp;atm/(mol&amp;nbsp;K))(323&amp;nbsp;K)Comparing the moles of NH3 and O2, we find that the moles of NH3 are multiplied by 5/4 to balance the equation.Now, let&#039;s determine the moles of water formed. From the balanced equation, we see that 4 moles of NH3 yield 6 moles of H2O.Therefore, the number of moles of H2O formed can be calculated as:nH2O=64×nNH3Now, we can calculate the mass of H2O formed using the molar mass of water, which is approximately 18.015 g/mol.mH2O=nH2O×molar&amp;nbsp;mass&amp;nbsp;of&amp;nbsp;H2OFinally, we substitute the calculated values and solve for the mass of water formed.

Determine the mass of water formed when 12.5 L NH3

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