What is the minimum amount of 6.9 M H2SO4 necessary to produce 27.8 g of H2 (g) according to the following reaction? 2Al(s)+3H2SO4(aq)→Al2(SO4)3(aq)+3H2(g)

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Alleei Alleei · Answer 1 · 2020-03-04T05:30:26+01:00

Answer : The minimum amount of 6.9 M [tex]H_2SO_4[/tex] needed is, 2.0 L

Explanation :

The given chemical reaction is:

[tex]2Al(s)+3H_2SO_4(aq)\rightarrow Al_2(SO_4)_3(aq)+3H_2(g)[/tex]

First we have to calculate the moles of [tex]H_2[/tex]

[tex]\text{Moles of }H_2=\frac{\text{Mass of }H_2}{\text{Molar mass of }H_2}[/tex]

Molar mass of [tex]H_2[/tex] = 2 g/mol

[tex]\text{Moles of }H_2=\frac{27.8g}{2g/mol}=13.9mol[/tex]

Now we have to calculate the moles of [tex]H_2SO_4[/tex]

From the balanced chemical reaction we conclude that,

As, 3 moles of [tex]H_2[/tex] produced from 3 moles of [tex]H_2SO_4[/tex]

So, 13.9 moles of [tex]H_2[/tex] produced from 13.9 moles of [tex]H_2SO_4[/tex]

Now we have to calculate the mass of [tex]H_2SO_4[/tex]

[tex]\text{Concentration of }H_2SO_4=\frac{\text{Moles of }H_2SO_4}{\text{Volume of }H_2SO_4}[/tex]

[tex]6.9M=\frac{13.9mol}{\text{Volume of }H_2SO_4}[/tex]

[tex]\text{Volume of }H_2SO_4=2.0L[/tex]

Thus, the minimum amount of 6.9 M [tex]H_2SO_4[/tex] needed is, 2.0 L