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enyata [817]
3 years ago
11

Nitrogen gas can be prepared by passing gaseous ammonia over solid copper (II) oxide at high temperatures. If 18.1 g of Nh3 is r

eacted with 90.4 g of CuO, which is the limiting reactant? How many grams of N2 will be formed? Explain how you solved for your answers.
Balanced Equation: 2NH3 + 3CuO → 3Cu + N2 + 3H2O
Chemistry
1 answer:
Scilla [17]3 years ago
6 0

I first converted the given grams of the reactants into moles, and then divided the moles by the coefficients in front of each of the reactant. The result with the smallest value will be the limiting reactant, and the value of CuO was the smallest, so it's the limiting reactant.

After figuring out which reactant is the limiting one, I took their given grams and converted it into moles, the divided it by the ratio of N2 to CuO (it's in the equation) to obtain the moles of N2, and then multiply it with the molar mass of N2 to get its mass in grams.

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What mass of the protein gelatin is needed to make 0.5 L of a 3 g/L gelatin solution? Show your work.
REY [17]

Answer:

m = 1.5 gram

Explanation:

Given that,

Density of protein gelatin, d = 3 g/L

The volume of protein gelatin, V = 0.5 L

We need to find the mass of the protein gelatin. The density of an object is given by :

d = m/V

Where

m is mass

m=d\times V\\\\m=3\ g/L\times 0.5\ L\\\\m=1.5\ g

So, the required mass is 1.5 gram.

6 0
3 years ago
PLEASE HELP!<br><br> See picture
Darya [45]

Answer:

See Explanation

Explanation:

The equation of the reaction;

KHSO4(aq) + KOH(aq) -------> K2SO4(aq) + H2O(l)

Number of moles of KHSO4 = 49.6 g/136.169 g/mol = 0.36 moles

Since the reaction is in a mole ratio of 1:1, 0.36 moles of K2SO4 is produced.

Number of moles of KOH = 25.3 g/56.1056 g/mol = 0.45 moles

Since the reaction is 1:1, 0.45 moles of K2SO4 is produced

Hence K2SO4 is the limiting reactant.

Mass of K2SO4 formed = 0.36 moles of K2SO4 * 174.26 g/mol = 62.7 g

So;

1 mole of KHSO4 reacts with 1 mole of KOH

0.36 moles of KHSO4 reacts with 0.36 * 1/1 = 0.36 moles of KOH

Amount of excess KOH = 0.45 moles - 0.36 moles = 0.09 moles

Mass of excess KOH = 0.09 moles * 56.1056 g/mol  = 5 g of excess KOH

5 0
3 years ago
Give the symbol for the element with the following orbital diagram:
wel

Answer: 0%

Explanation:

You got a bad grade

5 0
3 years ago
How many grams of material is lost in the aqueous phase if two extractions are carried out on 100 mL of a 5% (m/v) aqueous solut
san4es73 [151]

Answer:

4.94g of material

Explanation:

Partition coefficient (Kp) of a substance is defined as the ratio between concentration of organic solution and aqueous solution, that is:

Kp = <em>8 = Concentration in Ethyl acetate / Concentration in water</em>

100mL of a 5% solution contains 5g of material in 100mL of water. Thus:

8 = X / 100mL / (5g-X) / 100mL

<em>Where X is the amount of material in grams that comes to the organic phase.</em>

8 = X / 100mL / (5g-X) / 100mL

8 = 100X / (500-100X)

4000 - 800X = 100X

4000 = 900X

4.44g = X

<em>Thus, in the first extraction you will lost 4.44g of material from the aqueous phase.</em>

And will remain 5g-4.44g = 0.56g.

In the second extraction:

8 = X / 100mL / (0.56g-X) / 100mL

8 = 100X / (56-100X)

448 - 800X = 100X

448 = 900X

0.50g = X

<em>In the second extraction, you will extract 0.50g of material</em>

Thus, after the two extraction you will lost:

4.44g + 0.50g = <em>4.94g of material</em>

<em></em>

6 0
3 years ago
Will give brainliest
madreJ [45]

Answer:

C

Explanation:

The concept behind, is mole ratio of Al:FeO

5 0
3 years ago
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