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vladimir2022 [97]
3 years ago
15

If a reaction vessel initially contains 5 mol S and 10 mil O2, how many moles of S will be in the reaction vessel once the react

ants have reacted as much as possible
Chemistry
1 answer:
Alex_Xolod [135]3 years ago
6 0

Answer:

Explanation:

2S       +       3O₂       =       2SO₃

2moles      3 moles

2 moles of S react with 3 moles of O₂

5 moles of S will react with 3 x 5 / 2 moles of O₂

= 7.5 moles of O₂ .

O₂ remaining unreacted = 10 - 7.5 = 2.5 moles .

All the moles of S will exhausted in the reaction and 2.5 moles of oxygen will be left .  

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m_{Zn(OH)_2}=38.4g

Explanation:

Hello!

In this case, for the undergoing chemical reaction:

ZnO(s)+H_2O(l)\rightarrow Zn(OH)_2

We evaluate the yielded moles of zinc hydroxide by each reactant as shown below:

n_{Zn(OH)_2}^{by ZnO}=35.4gZnO*\frac{1molZnO}{81.38gZnO}*\frac{1molZn(OH)_2}{1molZnO}  =0.435molZn(OH)_2\\\\n_{Zn(OH)_2}^{by H_2O}=6.96gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{1molZn(OH)_2}{1molH_2O}  =0.386molZn(OH)_2

In such a way, since the water yields a smaller amount of zinc hydroxide we conclude it is the limiting reactant so the maximum mass is computed below:

m_{Zn(OH)_2}=0.386molZn(OH)_2*\frac{99.424 gZn(OH)_2}{1molZn(OH)_2} \\\\m_{Zn(OH)_2}=38.4g

Because the water limits the yielded amount of zinc hydroxide.

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