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4 years ago

Consider the balanced equation for the following reaction:

Chemistry

1 answer:

mafiozo [28]4 years ago

3 0

Answer:

There will remain 78.9 grams of NH3.

Explanation:

Step 1: Data given

Mass CuO = 17.3 grams

Molar mass CuO = 79.545 g/mol

Mass NH3 = 81.4 grams

Molar mass NH3 = 17.03 g/mol

Step 2: The balanced equation

3CuO(s) + 2NH3(g) → 3H2O(l) + 3Cu(s) + N2(g)

Step 3: Calculate moles CuO

Moles CuO = mass CuO/ molar mass CuO

Moles CuO = 17.3 grams / 79.545 g/mol

Moles CuO = 0.217 moles

Step 4: Calculate moles NH3

Moles NH3 = 81.4 grams / 17.03 g/mol

Moles NH3 = 4.78 moles

Step 5: Calculate the limiting reactant

CuO is the limiting reactant. It will completely react. NH3 is in excess. There will react 2/3 * 0.217 = 0.145 moles. There will remain 4.78 - 0.145 = 4.635 moles NH3

Step 6: Calculate mass NH3 remaining

Mass NH3 = 4.635 moles * 17.03 g/mol

Mass NH3 = 78.9 grams

There will remain 78.9 grams of NH3.

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A multistep reaction can only occur as fast as its slowest step. Therefore, it is the rate law of the slow step that determines

icang [17]

Answer:

b) rate = kAB.

Explanation:

Hello,

In this case, considering the given statement, we can notice that the rate law of the overall reaction will be determined for the slowest step, that is:

$A+B \rightarrow AB\ \ (slow)$

In such a way, we can infer that the rate law will contain both the concentration of A and B to the first power both, since their stoichiometric coefficients in the chemical equation are both one:

$rate=k[A][B]$

Thereby, answer is b) rate = kAB, that should be better rate = k[A][B] by expressing the concentrations.

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8 0

3 years ago

An object starts from rest and exhibits the motion represented by the graph. The object’s average velocity during the 3 second t

Rufina [12.5K]

Answer:

8.4 m/s

Explanation:

Initial velocity of the object is 0 as it was at rest.

We need to find the object’s average velocity during the 3 second to 5 second.

We can see from the graph that from 3 s to 5 s, the graph is a straight line. If we take the slope we can find the average velocity during the 3 second to 5 second.

Displacement between 3 s to 5 s is :

D = 24.8 m - 8 m = 16.8 m

Time = 5 s - 3 s = 2 s

Now,

Average velocity,

$v=\dfrac{\Delta y}{\Delta x}\\\\y=\dfrac{16.8\ m}{2\ m}\\\\y=8.4\ m$