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expeople1 [14]
2 years ago
13

Ammonia reacts with hypobromite OBr- as follows: NH3 OBr- ---> N2 Br- H20 which is the limiting reactant if 5.00 ml of 0.623

M NaOBr solution are added to
Chemistry
1 answer:
elixir [45]2 years ago
5 0

Answer:

Limiting reactant: NaOBr or OBr⁻.

Left over: 0.0006317 moles of ammonia remain unreacted.

Explanation:

Hello!

In this case, for the balanced chemical reaction:

2NH_3 +3OBr^- \rightarrow  N_2+ 3Br^- 3H_2O

Thus, when 5.00 ml of 0.623 M NaOBr react with 183 μL of 14.8 M NH3, in order to identify the limiting reactant, we first compute the moles of each reactant:

n_{OBr^-}=5.00mL*\frac{1L}{1000mL}*0.623\frac{molNaOBr}{L}*\frac{1molOBr^-}{1molNaOBr}=0.003115mol   OBr^-\\\\n_{NH_3}=183\mu L*\frac{1L}{1x10^6\mu L} *14.8\frac{molNH_3}{1L} =0.0027084molNH_3

Now, we can compute the moles of any product, say nitrogen gas, produced by each reactant:

n_{N_2}^{by\ NH_3} =0.0027084molNH_3*\frac{1molN_2}{2molNH_3} =0.0013542molN_2\\\\ n_{N_2}^{by\ OBr^-} =0.003115molOBr^-*\frac{1molN_2}{3molOBr^-} = 0.0010383molN_2

Therefore, since the hypobromite yields the fewest moles of nitrogen, we infer it is the limiting reactant. Moreover, for the left over of ammonia, we compute the required moles to yield 0.0010383 moles of nitrogen:

n_{NH_3}^{required} =0.0010383molN_2*\frac{2molNH_3}{1molN_2}  =0.002077molNH_3

So the left over is:

n_{NH_3}^{left\ over} = 0.0027084mol-0.0020767mol\\\\n_{NH_3}^{left\ over} =0.0006317molNH_3

Best regards!

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<h3>Further explanation</h3>

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Solution

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\tt \dfrac{V_1}{V_2}=\dfrac{n_1}{n_2}

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