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3 years ago

A sample of 8.5 g nh3 on oxidation produces 4.5 g of no. calculate the percent yield. reaction: 4 nh3 5 o2 → 4 no 6 h2o

1 answer:

5 0

30.% Determine the molar mass of NH3 and NO Atomic weight nitrogen = 14.0067 Atomic weight hydrogen = 1.00794 Atomic weight oxygen = 15.999 Molar mass NH3 = 14.0067 + 3 * 1.00794 = 17.03052 g/mol Molar mass NO = 14.0067 + 15.999 = 30.0057 g/mol Moles NH3 = 8.5 / 17.03052 = 0.499103962 mol Moles NO = 4.5 / 30.0057 = 0.149971505 mol Looking at the balanced equation, for every mole of NH3 consumed, you should get one mole of NO. So if we had 100% yield, we should have 0.499103962 moles of NO. But we don't. The percent yield is a simple matter of division by what we did get by what we should get. So 0.149971505 / 0.499103962 = 0.300481497 = 30.0481497% Rounding to 2 significant figures gives 30.% yield.

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3 years ago

An acid solution is 0.100 M in HCl and 0.210 M in H2SO4. What volume of a 0.150 M solution of KOH must be added to 500.0 mL of t

Dmitry_Shevchenko [17]

Answer:

When we add 0.33L of KOH only the HCl solution will be neutralized. When we add 1.4 L of KOH, all of the acid (the HCl and H2SO4 solution) will be neutralized.

Explanation:

Step 1: Data given

The solution has 0.100 M HCl and 0.210 M H2SO4

Molarity KOH = 0.150 M

Volume of acid solution = 500 mL = 0.5 L

Step 2: Calculate moles of HCl

Moles HCl = Molarity HCl * volume

Moles HCl = 0.100 M * 0.5 L

Moles HCl = 0.05 moles

Step 3: Calculate moles of H2SO4

Moles H2SO4 = 0.210 M * 0.5 L

Moles H2SO4 = 0.105 moles

Step 4: The balanced neutralization reaction of KOH with HCl can be written as:

KOH + HCl → KCl + H2O

The mole ratio is 1:1

This means to neutralize 0.05 moles HCl, we need 0.05 moles KOH

Step 5: The balanced neutralization reaction of KOH with H2SO4 can be written as:

2KOH + H2SO4 → K2SO4 + 2H2O

The mole ratio KOH: H2SO4 is 2:1

This means to neutralize 0.105 moles H2SO4 we need 0.210 moles KOH

Step 6: Calculate volume of KOH needed to neutralize the solution

To neutralize the HCl solution: 0.05 moles / 0.150 M = 0.33 L KOH needed

To neutralize the H2SO4 solution: 0.210 moles / 0.150 M = 1.4 L KOH needed

When we add 0.33L of KOH the HCl solution will be neutralized. When we add 1.4 L of KOH the HCl and H2SO4 solution will be neutralized.

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2 years ago