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Hydrogen gas is produced when zinc reacts with hydrochloric acid. If the actual yield of this reaction is 85%?

how many grams of zinc are needed to produce 112 l of H2 at STP?

a. 95 g
b. 180
c. 280
d. 380

User Mgutz
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1 Answer

5 votes

Answer:

d. 380g

Step-by-step explanation:

The reaction of production of hydrogen from zinc and HCl is:

Zn + 2HCl → ZnCl₂ + H₂(g)

At STP (1atm of pressure and 273.15K), 112L are:

n = PV /RT

n = 1atm×112L / 0.082atmL/molK×273.15K

n = 5.00 moles

That means you need to produce 5.00 moles of hydrogen. Based on the reaction, 1 mole of Zn produce 1 mole of H₂(g), as yield of reaction is 85%:

5.00 moles H₂(g) ≡ 5.00 moles of Zn / 85% = 5.88 moles of Zn you need

As molar mass of Zn is 65.38 g/mol:

5.88 moles Zn × (65.38g /mol) = 384 g of Zn ≡ d. 380g

I hope it helps!

User Winsmith
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