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Calculate the Experimental Molar Volume in L/mol of the Hydrogen gas, H2, if the volume of H2 at STP is 52.8 mL and the mass of Magnesium metal, Mg, used in the experiment is 0.055 g.

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Answer:

The Experimental Molar Volume in L/mol of the Hydrogen gas=23.36L/mol

Step-by-step explanation:

We are given that

Volume of H2 at STP=52.8mL

Mass of magnesium metal ,M(Mg)=0.055g

We have to find the Experimental Molar Volume in L/mol of the Hydrogen gas.

Molar mass of Mg=24.305 g/mol

Number of moles=
(given\;mass)/(molar\;mass)

Using the formula

Number of moles of Mg=
(0.055)/(24.305)moles

Number of moles of Mg=0.00226moles

Number of moles of Mg=Number of moles of H2

Number of moles of H2=0.00226moles

Molar volume of Hydrogen gas (H2)=
(volume\;at\;STP)/(No\;of\;moles\;H_2)

Molar volume of Hydrogen gas (H2)=
(52.8)/(0.00226)mL/mol

Molar volume of Hydrogen gas (H2)=
(52.8)/(0.00226)* 10^(-3)L/mol


1L=1000mL

Molar volume of Hydrogen gas (H2)=23.36L/mol

Hence, the Experimental Molar Volume in L/mol of the Hydrogen gas=23.36L/mol

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