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maxonik [38]
3 years ago
11

For the reaction N2(g) + 2H2(g) → N2H4(l), if the percent yield for this reaction is 100.0%, what is the actual mass of hydrazin

e (N2H4) produced when 59.20 g of nitrogen reacts with 6.750 g of hydrogen?
a. Molar mass of N2 = 28.01 g/mol
b. Molar mass of H2 = 2.016 g/mol
c. Molar mass of N2H4 = 32.05 g/mol
Chemistry
1 answer:
barxatty [35]3 years ago
4 0

Answer:

53.6 g of N₂H₄

Explanation:

The begining is in the reaction:

N₂(g) + 2H₂(g) → N₂H₄(l)

We determine the moles of each reactant:

59.20 g / 28.01 g/mol = 2.11 moles of nitrogen

6.750 g / 2.016 g/mol = 3.35 moles of H₂

1 mol of N₂ react to 2 moles of H₂

Our 2.11 moles of N₂ may react to (2.11 . 2) /1 = 4.22 moles of H₂, but we only have 3.35 moles. The hydrogen is the limiting reactant.

2 moles of H₂ produce at 100 % yield, 1 mol of hydrazine

Then, 3.35 moles, may produce (3.35 . 1)/2 = 1.67 moles of N₂H₄

Let's convert the moles to mass:

1.67 mol . 32.05 g/mol = 53.6 g

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Answer:

a. the mole fraction of CO in the mixture of CO and O2.

mole fraction = moles of CO/ Total moles of the mixture

Mole fraction of CO = 10/(10+12.5)=0.444

b. Reaction - CO(g)+½O2(g)→CO2(g)

Stoichiometry: 1 mole of CO react with 0.5mole of O2 to give 1 mole of CO2

So given,

At a certain point in the heating, 3.0 mol CO2 is present. Determine the mole fraction of CO in the new mixture.

3mol of CO2 is produced from 3 mols of CO and 1.5mol of O2

This means that unused mols are : 7mols of CO and 11mols of O2

Total product mixture = 3 + 7 + 11 = 21mols

mole fraction of CO = 7/21 = 0.33

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Answer:

hydrogen oxygen or nitrogen

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horrorfan [7]
The correct answer is E10.
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Which is NOT a physical property?
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Can Magnesium hydroxide be used as the base for this titration? Why or why not? 5. Describe the procedure that you will follow t
kvasek [131]

Explanation:

A.

Yes

B.

2KHP + Mg(OH)2 = 2KOH + Mg(HP)2

KHP is an acidic salt which reacts with Mg(OH)2 which is a base via double displacement/decomposition reaction.

C.

• One approach is to prepare the solution volumetrically using KHP crystals.

• Preparing 1 liter of 0.1 M KHP you would add 0.1 moles of KHP

Molar mass of KHP(C8H5KO4)

= (12*8) + (5*1) + 39 + (16*4)

= 204 g/mol

Number of moles = mass/molar mass

= 20.4/204

= 0.1 mol.

• This is added to a 1 liter volumetric flask, add deionized water until near the fill line, stopper and mix

D.

Number of moles = molar concentration * volume

= 0.025 * 0.100

= 0.0025 mol

Equation for the reaction:

Ba(OH)2 + 2KHP --> Ba(KP)2 + 2H2O

1 mole of Ba(OH)2 reacted with 2 moles of KHP. By stoichiometry,

Number of moles of Ba(OH)2 = 0.0025/2

= 0.00125 mol.

Molarity = number of moles/volume

= 0.00125/0.01

= 0.125 M of Ba(OH)2.

4 0
3 years ago
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