1.22 g H is allowed to react with 10.2 g N2. producing 1.48 g NH dustrial gas re

Question

1.22 g H is allowed to react with 10.2 g N2. producing 1.48 g NH dustrial gas reacts to the Part A What is the theoretical yield in grams for this reaction under the given conditions? Express your answer to three significant figures and include the appropriate units. cess has a uticals , resulting itn mical Vaue Units Submlt Hints My Answers Give Up Review Part Part B What is the percent yield for this reaction under the glven conditions? Express your answer to three significant figures and include the appropriate units. Value Units Submit Hints My Answers Give Up Review Parn

Explanation / Answer

A) N2 + H2 ---> NH3

balence the reaction N2 +3H2 --> 2NH3

one mole of N2 react with 3 mole of H2 produce 2 mole of NH3

moles of N2 = (10.2 g/28.0 g) = 0.364 moles

moles of H2 = (1.22g/2.0g) = 0.61 moles

reacting ratio is 1:3:2 (N2:H2:NH3)

0.61 moles of H2 react with 0.2033 moles of N2 to produce 0.305 moles of NH3

mass of NH3 theoritical yeild = (0.305 moles x 17.031 g/mol ) = 5.1944 g

B) percentage yeild = actual yeild / theoritical yeild

moles of NH3 actual yeild = (1.48g / 17.031g/mol) = 0.0869 moles

percentage yeild =  (0.0869 moles / 0.305 moles) x 100 = 28.49 %

2) B 29.33g C6H12O6 in 1.33 L solution

molarity of solution = (moles of solute / volume of solution )

moles of C6H12O6 = (29.33g /180.1559 g/mol) = 0.1628 moles

molarity of solution (M) =  0.1628 moles / 1.33 L = 0.1224 M

c) moles of NaCl = (38.3mg / 58.44 g/mol) = 0.65537 m mol

molarity of solution (M) = 0.65537 m mol / 123.0ml = 0.005328 M

3) H2O molecules

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