Magnesium metal reacts with HCl solution, liberatingH 2 gas and generating Mg 2+

Question

Magnesium metal reacts with HCl solution, liberatingH2 gas and generating Mg2+ cations insolution. A 2.110-g sample of Mg metal is added to 150.0 mL of a4.00 M HCl solution, and the reaction goes to completion.the massof H2 formed is 0.174 g

      Mg +2H+  ---> Mg2+ + H2 stoichiometry number: 1 2 moles of reactants: Mg ( 2.110/24.3 =0.087 mol; HCl (4.00 x 0.15 = 0.6 mol) Because 0.087 x 2 = 0.174 mol < 0.6 x 1 =0.6 mol, HCl is in excess. Then the moles of H2 can be calculated usingthe moles of Mg. The moles of H2 = the moles of Mg metal = 0.087mol; FW of H2 = 2 g/mol mass of H2 = 0.087 mol x 2 g/mol = 0.174 g

What is the concentration of Mg2+ ionsin the solution? Magnesium metal reacts with HCl solution, liberatingH2 gas and generating Mg2+ cations insolution. A 2.110-g sample of Mg metal is added to 150.0 mL of a4.00 M HCl solution, and the reaction goes to completion.the massof H2 formed is 0.174 g

      Mg +2H+  ---> Mg2+ + H2 stoichiometry number: 1 2 moles of reactants: Mg ( 2.110/24.3 =0.087 mol; HCl (4.00 x 0.15 = 0.6 mol) Because 0.087 x 2 = 0.174 mol < 0.6 x 1 =0.6 mol, HCl is in excess. Then the moles of H2 can be calculated usingthe moles of Mg. The moles of H2 = the moles of Mg metal = 0.087mol; FW of H2 = 2 g/mol mass of H2 = 0.087 mol x 2 g/mol = 0.174 g

What is the concentration of Mg2+ ionsin the solution? stoichiometry number: 1 2 moles of reactants: Mg ( 2.110/24.3 =0.087 mol; HCl (4.00 x 0.15 = 0.6 mol) Because 0.087 x 2 = 0.174 mol < 0.6 x 1 =0.6 mol, HCl is in excess. Then the moles of H2 can be calculated usingthe moles of Mg. The moles of H2 = the moles of Mg metal = 0.087mol; FW of H2 = 2 g/mol mass of H2 = 0.087 mol x 2 g/mol = 0.174 g

What is the concentration of Mg2+ ionsin the solution?

Explanation / Answer

Considering the moles of Mg2+ ions in solution should beequivalent to the amount of Mg metal that reacted, you then take the number of moles of Mg2+ and thendivided by your total volume (in L) to obtain molarity, which wouldbe your measure of concentration. So in this case, take 0.087 moles of Mg = 0.087 moles ofMg2+ 0.087 moles Mg2+ / 0.150 L solution = Molarity ofMg2+ --> Which is your concentration

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.