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Lemur [1.5K]
3 years ago
5

A 12.2-g sample of X reacts with a sample of Y to form 78.9g of XY. what mass of Y reacted?​

Chemistry
1 answer:
barxatty [35]3 years ago
3 0

Answer:

The answer to your question is:  y = 66.7g

Explanation:

To solve this problem we must remember the Lavoisier law of conservation of mass, that stated that the mass can not be created or destroyed but it changes forms.

So, the mass of the reactants is equal to the mass of the products.

Process

                x          +          y             ⇒           xy

             12.2g                   ?                         78.9 g

           12.2 g + y = 78.9g

           y = 78.9 - 12.2

           y = 66.7g

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krek1111 [17]

The answer is 308 K.


The formula is C + 273.15 = K

this meaning the formula for this problem would be 35 + 273.15 = 308.15.

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Answer:

6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.

Explanation:

We are given the chemical equation:

\displaystyle 2\text{NH$_3$}_\text{(g)} + 3\text{CuO}_\text{(s)} \longrightarrow \text{N$_2$}_\text{(g)}  + 3\text{Cu}_\text{(s)}+3\text{H$_2$O}_\text{(g)}

And we want to determine the amount of products produced when 12.5 moles of NH₃ is reacted with excess CuO.

Compute using stoichiometry. From the equation, we can see the following stoichiometric ratios:

  • The ratio between NH₃ and N₂ is 2:1. (i.e. One mole of N₂ is produced from every two moles of NH₃.)
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  • The ratio between NH₃ and H₂O is 2:3. (i.e. Three moles of H₂O or Cu is produced frome every two moles of NH₃.)

Dimensional Analysis:

  • The amount of N₂ produced:

\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{1\text{ mol N$_2$}}{2\text{ mol NH$_3$}} = 6.25\text{ mol N$_2$}

  • The amount of Cu produced:

\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{3\text{ mol Cu}}{2\text{ mol NH$_3$}} = 18.8\text{ mol Cu}

  • And the amount of H₂O produced:

\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{3\text{ mol H$_2$O}}{2\text{ mol NH$_3$}} = 18.8\text{ mol H$_2$O}

In conclusion, 6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.

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