Questions: Consider the reaction Mg(s) + HCl(ag) -> MgCl2(ag) + H2(g) a) Balance the reaction 1 Mg(s) + 2 HCl(ag) -> 1 MgCl2(ag) + 1 H2(g) b) A reaction contains 2.50 g of solid magnesium is mixed with 2.50 mL of 0.100 M HCl. Determine the limiting reagent, and the excess reagent. 2.50 mL HCl x (1 L HCl / 1000 mLHCl) = 0.0025 L -> 0.100 M = (mlec HCl / 0.0025) = 0.00025 molHCl x (1 md MgCl / 2 md HCl) = 0.000125 mol 0.000125 mol MgCl2 HCl is L.R c) How many grams of MgCl2 can be produced from the above reaction based on the 0.000125 molM MgCl2 x (95.2112 MgCl2 / 1 mal MgCl2) = 0.0119 MgCl2 d) How many grams of hydrogen gas will be evolved from the reaction based on the limiting reagent? 0.000125 md MgCl2 x (1 mol H2 / 1 mol MgCl) x (2.016 g H2 / 1 molH) = 0.000252 g e) How many grams of the excess reagent left unreacted? 0.100 M = (moles of HCl / 0.00250 L) = moles of HCl = 2.5 x 10^-4 moles 2.5 x 10^-4 mol HCl x (36.461, HCl / 1 mol) = 9.12 x 10^-3 g HCl

Consider the reaction Mg(s) + HCl(ag) -> MgCl2(ag) + H2(g) a) Balance the reaction 1 Mg(s) + 2 HCl(ag) -> 1 MgCl2(ag) + 1 H2(g) b) A reaction contains 2.50 g of solid magnesium is mixed with 2.50 mL of 0.100 M HCl. Determine the limiting reagent, and the excess reagent. 2.50 mL HCl x (1 L HCl / 1000 mLHCl) = 0.0025 L -> 0.100 M = (mlec HCl / 0.0025) = 0.00025 molHCl x (1 md MgCl / 2 md HCl) = 0.000125 mol 0.000125 mol MgCl2 HCl is L.R c) How many grams of MgCl2 can be produced from the above reaction based on the 0.000125 molM MgCl2 x (95.2112 MgCl2 / 1 mal MgCl2) = 0.0119 MgCl2 d) How many grams of hydrogen gas will be evolved from the reaction based on the limiting reagent? 0.000125 md MgCl2 x (1 mol H2 / 1 mol MgCl) x (2.016 g H2 / 1 molH) = 0.000252 g e) How many grams of the excess reagent left unreacted? 0.100 M = (moles of HCl / 0.00250 L) = moles of HCl = 2.5 x 10^-4 moles 2.5 x 10^-4 mol HCl x (36.461, HCl / 1 mol) = 9.12 x 10^-3 g HCl
Transcript text: Consider the reaction \[ \mathrm{Mg}(\mathrm{~s})+\mathrm{HCl}(\mathrm{ag}) \rightarrow \mathrm{MgCl}_{2}(\mathrm{ag})+\mathrm{H}_{2}(\mathrm{~g}) \] a) Balance the reaction \[ 1 M_{2(s)}+2 H C l_{(4)} \rightarrow 1 M_{g} C l_{2(y)}+1 \mathrm{H}_{2(g)} \] b) A reaction contains 2.50 g of solid magnesium is mixed with 2.50 mL of 0.100 M HCl. Determine the limiting reagent, and the excess reagent. \[ \begin{array}{l} \begin{array}{r} 2.50 \mathrm{~mL} \mathrm{HCl} \times \frac{1 \mathrm{~L} \mathrm{HCl}}{1000 \mathrm{mLHCl}}=0.0025 \mathrm{~L} \rightarrow 0.100 \mathrm{M}=\frac{\text { mlec HCl }}{0.0025}=0.00025 \mathrm{molHCl} \times \frac{1 \mathrm{md} \mathrm{MyCl}}{2 \mathrm{md} \mathrm{HCl}}= \\ 0.000125 \mathrm{~mol} \end{array} \\ 0.000125 \mathrm{~mol} \mathrm{mgl}_{2} \mathrm{HCl} \text { is L.R } \\ \text { c) How many grams of } \mathrm{MgCl}_{2} \text { can be produced from the above reaction based on the } \end{array} \] \[ 0.000125 \mathrm{molM} \mathrm{M}_{2} \mathrm{Cl}_{2} \times \frac{95.2112 \mathrm{M}_{2} \mathrm{Cl}_{2}}{1 \mathrm{mal} \mathrm{M}_{2} \mathrm{Cl}_{2}}=0.0119 \mathrm{M}_{2} \mathrm{M}_{2} \mathrm{Cl}_{2} \] d) How many grams of hydrogen gas will be evolved from the reaction based on the limiting reagent? \[ 0.000125 \mathrm{md} \mathrm{MeCl} 2 \times \frac{1 \mathrm{~mol} \mathrm{H2}}{1 \mathrm{~mol} \mathrm{MgCl}} \times \frac{2.016 \mathrm{~g} \mathrm{H}_{2}}{1 \mathrm{molH}}=0.000252 \mathrm{~g} \] \[ \begin{array}{l} \text { e) How many grams of the excess reagent left unreacted? } \\ 0.100 \mathrm{M}=\frac{\text { moles of } \mathrm{HCl}}{0.00250 \mathrm{~L}}=\text { moles of } \mathrm{HCl}=2.5 \times 10^{-4} \text { moles } \\ 2.5 \times 10^{-4} \mathrm{~mol} \mathrm{HCl} \times \frac{36.461, \mathrm{HCl}}{1 \mathrm{~mol}}=9.12 \times 10^{-3} \mathrm{~g} \mathrm{HCl} \\ \end{array} \]
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Solution

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Solution Steps

Step 1: Balance the Reaction

The given reaction is: \[ \mathrm{Mg}(\mathrm{s}) + \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{MgCl}_{2}(\mathrm{aq}) + \mathrm{H}_{2}(\mathrm{g}) \]

To balance the reaction, we need to ensure that the number of atoms of each element is the same on both sides of the equation. The balanced equation is: \[ \mathrm{Mg}(\mathrm{s}) + 2\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{MgCl}_{2}(\mathrm{aq}) + \mathrm{H}_{2}(\mathrm{g}) \]

Step 2: Determine the Limiting Reagent

We have 2.50 g of solid magnesium and 2.50 mL of 0.100 M HCl.

  1. Calculate moles of Mg: \[ \text{Molar mass of Mg} = 24.305 \, \text{g/mol} \] \[ \text{Moles of Mg} = \frac{2.50 \, \text{g}}{24.305 \, \text{g/mol}} = 0.1029 \, \text{mol} \]

  2. Calculate moles of HCl: \[ \text{Volume of HCl} = 2.50 \, \text{mL} = 0.00250 \, \text{L} \] \[ \text{Moles of HCl} = 0.100 \, \text{M} \times 0.00250 \, \text{L} = 0.00025 \, \text{mol} \]

  3. Determine the limiting reagent: The balanced equation shows that 1 mole of Mg reacts with 2 moles of HCl. Therefore, the moles of HCl required for 0.1029 moles of Mg is: \[ 0.1029 \, \text{mol Mg} \times 2 \, \text{mol HCl/mol Mg} = 0.2058 \, \text{mol HCl} \] Since we only have 0.00025 mol of HCl, HCl is the limiting reagent.

Step 3: Calculate Grams of MgCl\(_2\) Produced

Using the limiting reagent (HCl), calculate the amount of MgCl\(_2\) produced:

  1. Moles of MgCl\(_2\) produced: \[ \text{Moles of MgCl}_2 = \frac{0.00025 \, \text{mol HCl}}{2} = 0.000125 \, \text{mol MgCl}_2 \]

  2. Convert moles of MgCl\(_2\) to grams: \[ \text{Molar mass of MgCl}_2 = 95.211 \, \text{g/mol} \] \[ \text{Grams of MgCl}_2 = 0.000125 \, \text{mol} \times 95.211 \, \text{g/mol} = 0.0119 \, \text{g} \]

Final Answer

  • Balanced Reaction: \(\boxed{\mathrm{Mg}(\mathrm{s}) + 2\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{MgCl}_{2}(\mathrm{aq}) + \mathrm{H}_{2}(\mathrm{g})}\)
  • Limiting Reagent: \(\boxed{\text{HCl}}\)
  • Grams of MgCl\(_2\) Produced: \(\boxed{0.0119 \, \text{g}}\)
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