Question

You are performing a reaction involving the collection of a gas at SATP. If 5.00 g of magnesium is added to 100.0 cm3 of 0.100 mol dm−3 HCl, determine the volume of hydrogen gas that will be collected at SATP.

Mg(s)+2HCl(aq)→MgCl2(aq)+H2(g)

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

0.124 dm3

Explanation:

mol of Mg = 5.00g/24.31 g mol-1 = 0.206

mol of HCl = 100cm3/1000 x 0.100 mol dm-3 = 0.01

mol ratio Mg : HCl = 1 : 2 = 0.206 : 0.412

since 0.412 > 0.01, HCl is the limiting reagent

mol ratio HCl : H2 = 2 : 1

mol of H2 = 0.01/2 = 0.005

V = nRT/p

= (0.05)(8.31)(298K)/100kPa

= 0.124 dm3