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Work through the following five steps of the limiting reagent problem in your notebook. Make sure you use the proper units. Either scan or take a digital photo of your completed work, and upload it below. If you have no access to a scanner or digital camera or phone, you can type your responses in the text box below. Show your work! When hydrogen chloride reacts with magnesium, magnesium chloride and hydrogen gas are formed. Write a balanced equation for this reaction. If 10.1 g of magnesium react with 26.0 g of hydrogen chloride, what is the limiting reagent

Answer :

petegaffney
The balanced equation for the reaction between hydrogen chloride (HCl) and magnesium (Mg) is:

2HCl + Mg -> MgCl2 + H2

To find the limiting reagent, we need to calculate the amount of magnesium chloride and hydrogen gas produced from both reactants and see which reactant produces less product.

First, let’s find the molar masses:

• Mg: 24.31 g/mol
• HCl: 1.01 g/mol (hydrogen) + 35.45 g/mol (chlorine) = 36.46 g/mol

Now, let’s calculate the number of moles of each reactant:

• Moles of Mg = mass / molar mass = 10.1 g / 24.31 g/mol = 0.415 mol
• Moles of HCl = mass / molar mass = 26.0 g / 36.46 g/mol = 0.712 mol

Now, using the balanced equation, we can see that 1 mole of Mg reacts with 2 moles of HCl to produce 1 mole of MgCl2 and 1 mole of H2.

For Mg, if 0.415 moles of Mg are reacted, it requires 2 * 0.415 = 0.83 moles of HCl.
For HCl, if 0.712 moles of HCl are reacted, it requires 0.712 / 2 = 0.356 moles of Mg.

Since there is less Mg (0.415 moles) than what is required by HCl (0.83 moles), Mg is the limiting reagent.

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