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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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Si uno pone un hilo en un hielo y un poco de sal el hilo se levanta el hilo con el hielo. Por que sé relaciona con calor y tempe
777dan777 [17]

Answer:

Explanation:

El hielo encierra la primavera antes de espolvorear la sal.

Cuando se rocía sal sobre el hielo, disminuye el punto de fusión del hielo 32 ° F a un poco por debajo de 32 °, por lo tanto, se acumula.

A medida que el hielo se vuelve a congelar, encierra la primavera

3 0
3 years ago
3. A compound consists of 91.63 grams of carbon, 7.69 grams of hydrogen and 40.81 grams of
Vikentia [17]

Answer:

Molecular formula = C₁₂H₁₂O₄

Empirical formula is C₃H₃O.

Explanation:

Given data:

Mass of C = 91.63 g

Mass of H = 7.69 g

Mass pf O = 40.81 g

Molar mass of compound = 220 g/mol

Empirical formula = ?

Molecular formula = ?

Solution:

Number of gram atoms of H = 7.69 / 1.01 = 7.61

Number of gram atoms of O = 40.81 / 16 = 2.55

Number of gram atoms of C = 91.63 / 12 = 7.64

Atomic ratio:

            C                      :      H                :         O

           7.64/2.55          :    7.61 /2.55    :       2.55/2.55

               3                     :          3               :        1

C : H : O = 3 : 3 : 1

Empirical formula is C₃H₃O.

Molecular formula:

Molecular formula = n (empirical formula)

n = molar mass of compound / empirical formula mass

Empirical formula mass  = 3×12+ 3×1.01 +16 = 55.03

n = 220 / 55.03

n = 4

Molecular formula = 4 (empirical formula)

Molecular formula = 4 (C₃H₃O)

Molecular formula = C₁₂H₁₂O₄

8 0
3 years ago
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UNO [17]

Answer:

A) 0 charge

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neutron=neutral

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