Answer:
a. Copper (II) oxide is the limiting reactant.
b. [tex]m_{N_2}=5.30g[/tex]
c. [tex]Y=87\%[/tex]
d. [tex]m_{NH_3}=2.60gNH_3[/tex]
Explanation:
Hello,
In this case, for the given reaction:
[tex]2 NH_3(g) + 3 CuO(s) \rightarrow N_2(g) + 3 Cu(s) + 3 H_2O(l)[/tex]
a. The limiting reactant is identified by computing the available moles of ammonia and the moles of ammonia that react with 45.2 g of copper (I) oxide as shown below:
[tex]n_{NH_3}^{Available}=9.05gNH_3*\frac{1molNH_3}{17gNH_3}=0.532molNH_3\\n_{NH_3}^{Reacted}=45.2gCuO*\frac{1molCuO}{79.545gCuO}*\frac{2molCuO}{3molCuO} =0.379molNH_3[/tex]
In such a way, as there more ammonia available than that is reacted, we say it is in excess and the copper (II) oxide the limiting reactant.
b. Here, with the reacting moles of ammonia, we compute the yielded grams of nitrogen:
[tex]m_{N_2}=0.379molNH_3*\frac{1molN_2}{2molNH_3}*\frac{28gN_2}{1molN_2}\\m_{N_2}=5.30g[/tex]
c. Now, since the 5.30 g of nitrogen are the expected grams of it, the percent yield of nitrogen is compute by dividing the real obtained mass over the theoretical previously computed mass:
[tex]Y=\frac{4.61g}{5.30g} *100\%\\Y=87\%[/tex]
d. Finally, as 0.532 moles of ammonia are available, but just 0.379 moles react, the unreacted moles are:
[tex]n_{NH_3}=0.532molNH_3-0.379molNH_3=0.153molNH_3[/tex]
That in grams are:
[tex]m_{NH_3}=0.153molNH_3*\frac{17gNH_3}{1molNH_3}=2.60gNH_3[/tex]
Best regards.
The limiting reactant is CuO, the percent yield of nitrogen gas is 29%.
The equation of the reaction is;
2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l)
Number of moles of NH3 = 9.05 g/17 g/mol = 0.53 moles
Number of moles of CuO = 45.2 g/80 g/mol = 0.565 moles
Since 2 moles of NH3 reacts with 3 moles of CuO
0.53 moles of NH3 reacts with 0.53 moles × 3 moles/ 2 moles
= 0.795 moles
We can see that there is not enough CuO in the system hence it is the limiting reactant.
Number of moles of N2 produced = 0.565 moles × 28 g/mol =15.82 g of N2
Percent yield = Actual yield/Theoretical yield × 100/1
Percent yield = 4.61 g/15.82 g × 100/1
Percent yield = 29%
If 2 moles of NH3 reacts with 3 moles of CuO
x moles of NH3 reacts with 0.565 moles of CuO
x = 2 moles × 0.565 moles/3 moles
x = 0.38 moles
Number of moles of excess reactant left over = 0.53 moles - 0.38 moles
= 0.15 moles
Mass of excess reactant left over = 0.15 moles × 17 g/mol = 2.55 g
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