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inessss [21]
3 years ago
8

What mass of ammonia can be produced if 13.4 grams of nitrogen gas reacted ?

Chemistry
1 answer:
aivan3 [116]3 years ago
3 0

Answer:

If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia

Explanation:

Step 1: Data given

Mass of nitrogen gas (N2) = 13.4 grams

Molar mass of N2 = 28 g/mol

Molar mass of NH3 = 17.03 g/mol

Step 2: The balanced equation

N2 + 3H2 → 2NH3

Step 3: Calculate moles of N2

Moles N2 = Mass N2 / molar mass N2

Moles N2 = 13.4 grams / 28.00 g/mol

Moles N2 = 0.479 moles

Step 4: Calculate moles of NH3

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

For 0.479 moles N2 we'll produce 2*0.479 = 0.958 moles

Step 5: Calculate mass of NH3

Mass of NH3 = moles NH3 * molar mass NH3

Mass NH3 = 0.958 moles * 17.03 g/mol

Mass NH3 = 16.3 grams

If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia

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Delvig [45]

Answer:

94.325 g

Explanation:

We'll begin by converting 350 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

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350 mL = 0.35 L

Next, we shall determine the number of mole of KC₂H₃O₂ in the solution. This can be obtained as follow:

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Molarity of KC₂H₃O₂ = 2.75 M

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Molarity = mole /Volume

2.75 = Mole of KC₂H₃O₂ / 0.35

Cross multiply

Mole of KC₂H₃O₂ = 2.75 × 0.35

Mole of KC₂H₃O₂ = 0.9625 mole

Finally, we shall determine the mass of KC₂H₃O₂ needed to prepare the solution. This can be obtained as illustrated below:

Mole of KC₂H₃O₂ = 0.9625 mole

Molar mass of KC₂H₃O₂ = 39 + (12×2) +(3×1) + (16×2)

= 39 + 24 + 3 + 32

= 98 g/mol

Mass of KC₂H₃O₂ =?

Mass = mole × molar mass

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Mass of KC₂H₃O₂ = 94.325 g

Thus, the mass of KC₂H₃O₂ needed to prepare the solution is 94.325 g

6 0
2 years ago
25. 00 ml of a buffer solution contains 0. 500 m hclo and 0. 380 m naclo. if 50. 00 ml of water is added to the buffer, what are
geniusboy [140]

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Calculation of number of moles of each component,

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Number of moles of HClO = 0.0125 mole

Molarity of NaClO  = number of moles/volume in lit =  0. 38 M

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Number of moles of  NaClO  = 0.95 mole

Calculation of new concentration at volume 50 ml ( 0.05L)

Molarity of HClO = number of moles/volume in lit = 0.0125 mole/0.05L

Molarity of HClO = 0.25M

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Molarity of NaClO  = 19 M

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5 0
1 year ago
Write balanced equations for each of the following by inserting the correct coefficients in the blanks:
Aleksandr-060686 [28]

Balanced equation :

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Balancing a chemical equation :

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According to the law of conservation of mass, when a chemical reaction occurs, the mass of the products should be equal to the mass of the reactants. Therefore, the amount of the atoms in each element does not change in the chemical reaction. As a result, the chemical equation that shows the chemical reaction needs to be balanced. A balanced chemical equation occurs when the number of the atoms involved in the reactants side is equal to the number of atoms in the products side.

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3 0
1 year ago
___AsCl3+____H2S-->___As2S3+___HCI​
Alex17521 [72]

Answer:

2 AsCl₃ + 3 H₂S → As₂S₃ + 6 HCl

Explanation:

When we balance a chemical equation, what we are trying to do is to achieve the same number of atoms for each element on both sides of the arrow. On the right of the arrow is where we can find the products, while the reactants are found on the left of the arrow.

We usually balance O and H atoms last.

AsCl₃ + H₂S → As₂S₃ +HCl

<u>reactants</u>

As --- 1

Cl --- 3

H --- 2

S --- 1

<u>products</u>

As --- 2

Cl --- 1

H --- 1

S --- 3

2 AsCl₃ + H₂S → As₂S₃ +HCl

<u>reactants</u>

As --- 2

Cl --- 6

H --- 2

S --- 1

<u>products</u>

As --- 2

Cl --- 1

H --- 1

S --- 3

The number of As atoms is now balanced.

2 AsCl₃ + 3 H₂S → As₂S₃ +HCl

<u>reactants</u>

As --- 2

Cl --- 6

H --- 6

S --- 3

<u>products</u>

As --- 2

Cl --- 1

H --- 1

S --- 3

The number of S atoms is now equal on both sides.

2 AsCl₃ + 3 H₂S → As₂S₃ + 6 HCl

<u>reactants</u>

As --- 2

Cl --- 6

H --- 6

S --- 3

<u>products</u>

As --- 2

Cl --- 6

H --- 6

S --- 3

The equation is now balanced.

3 0
2 years ago
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