In the balanced redox reaction 2 Al(s) + 6 HCl(aq) → 2 AlCl3(aq) + 3 H2(g) 80.0 mL of 2.0 M HCl would (theoretically) react with how many grams of Al(s)? Assume that the reaction is quantitative.
A. 1.44 g
B. 4.32 g
C. 1,440 g
D. 2,030 g
E. None of the above

Answer :

The chemical reaction is written as:

 2 Al(s) + 6 HCl(aq) → 2 AlCl3(aq) + 3 H2(g)

We use the volume and the concentration of the acid to determine the number of moles HCl used. Then, we use relation from the reaction to determine the moles Al used. Finally, we convert it to grams by the molar mass.

(2.0 mol HCl / L) (.080 L) ( 2 mol Al / 6 mol HCl ) ( 26.98 g / mol ) = 1.44 g Al is used.

The grams of Al to react is mathematically given as

X = 1.44 g of Al  

What is the amount in grams of Al?

Question Parameters:

In the balanced redox reaction 2 Al(s) + 6 HCl(aq) → 2 AlCl3(aq) + 3 H2(g) 80.0 mL of 2.0 M HCl

Generally, the equation for the Chemical Reaction is mathematically given as

2 Al(s) + 6 HCl(aq) → 2 AlCl3(aq) + 3 H2(g)

Therefore, as volume and the concentration of the acid to decides the numeral of moles

X=(2.0 mol HCl / L) (0.080 L) ( 2 mol Al / 6 mol HCl ) ( 26.98 g / mol )

X = 1.44 g of Al  

Therefore

grams of Al(s) to react is 1.44 g of Al  

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