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.

Answers

Answer 1

Answer: The volume of hydrogen produced is 0.122 L

Explanation:

To calculate the moles :

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

[tex]\text{Moles of solute}={\text{Molarity}}\times {\text{Volume in L}}[/tex]  

[tex]\text{Moles of} Mg=\frac{5.00g}{24g/mol}=0.208moles[/tex]

[tex]\text{Moles of} HCl=[/tex][tex]0.100moldm^{-3}\times 0.1dm^3=0.01mol[/tex]

The balanced chemical equation is:

[tex]Mg+2HCl\rightarrow MgCl_2+H_2[/tex]  

According to stoichiometry :

2 moles of require = 1 mole of

Thus 0.01 moles of [tex]HCl[/tex] will require=[tex]\frac{1}{2}\times 0.01=0.005moles[/tex]  of [tex]Mg[/tex]

Thus is the limiting reagent as it limits the formation of product and is the excess reagent.

As 2 moles of give = 1 mole of

Thus 0.01 moles of [tex]HCl[/tex] give =[tex]\frac{1}{2}\times 0.01=0.005moles[/tex]  of [tex]H_2[/tex]

According to ideal gas equation:

[tex]PV=nRT[/tex]

P = pressure of gas = 1.0 atm  (SATP)

V = Volume of gas = ?  

n = number of moles = 0.005

R = gas constant =[tex]0.0821Latm/Kmol[/tex]

T =temperature =[tex]25^0C=(25+273)K=298K[/tex] (SATP)

[tex]V=\frac{nRT}{P}[/tex]

[tex]V=\frac{0.005mol\times 0.0820 L atm/K mol\times 298K}{1.0atm}=0.122L[/tex]

Thus volume of hydrogen produced is 0.122 L


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Z
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