Welcome to Westonci.ca, your one-stop destination for finding answers to all your questions. Join our expert community now! Connect with professionals ready to provide precise answers to your questions on our comprehensive Q&A platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
Answer: 0.276g of H2 could be obtained from the reaction, considering the mass of reactants given
Explanation:
The question requires us to determine the amount of hydrogen gas (H2) that would be obtained when 10.0g of metallic magnesium (Mg) and 10.0g of chloridric acid (HCl) are reacted.
The following unbalanced chemical equation was provided:
[tex]Mg+HCl\rightarrow MgCl_2+H_2[/tex]To solve this problem, we'll need:
1) obtain the balanced chemical equation;
2) determine the limiting reactant considering the amount of reactants used and the stoichiometry of the reaction;
3) calculate the mass of hydrogen gas produced, considering the limiting reactant
1) Balancing the chemical equation
From the unbalaced chemical equation, we can see that there is the same amount of Mg on both sides of the equation, but we need to adjust the amount of H and Cl atoms.
There are 2 Cl atoms and 2 H atoms on the right side, and 1 Cl and 1 H atom on the left side, thus we can adjust the coefficient of HCl from 1 to 2 to adjust these elements. The balanced chemical equation can be written as:
[tex]Mg+2HCl\rightarrow MgCl_2+H_2[/tex]2) Determining the limiting reactant
Now that we know the balanced chemical equation, we can say that 2 moles of HCl are necessary to react with 1 mol of Mg. Thus, we can write:
1 mol Mg ----------------------- 2 mol HCl
From this stoichiometric relation, we can calculate the amount of HCl that would be necessary to react with 10.0 g of Mg.
First, let's determine the amount of moles of Mg contained in 10.0g of this metal and the number of moles of HCl in 10.0g of this acid:
[tex]\begin{gathered} number\text{ of moles = }\frac{mass\text{ of sample \lparen g\rparen}}{molar\text{ mass \lparen g/mol\rparen}} \\ \\ number\text{ of moles of Mg = }\frac{10.0g}{24.31g/mol}=0.411mol\text{ of Mg} \\ \\ number\text{ of moles of HCl = }\frac{10.0g}{36.46g/mol}=0.274mol\text{ of HCl} \end{gathered}[/tex](the atomic mass of Mg is 24.31 amu, which is numerically identical to its molar mass, 24.31 g/mol, and the molar mass of HCl is 36.4g g/mol)
Therefore, 0.411 moles of Mg and 0.274 moles of HCl were used in the reaction.
Next, we can determine how many moles of HCl would be necessary to react with 0.411 moles of Mg:
1 mol Mg ----------------------- 2 mol HCl
0.411 mol Mg ----------------- x
Solving for x, we have that 0.822 moles of HCl would be necessary to react with the given amount of Mg. Since the amount actually used of HCl (0.274 mol) is smaller than the necessary amount, we can say that HCl is the limiting reactant.
3) Calculating the mass of H2 obtained
Using the limiting reactant and the stoichiometry of the reaction, we can determine the amount of H2 that could be produced.
From the balanced chemical equation, we know that 2 moles of HCl are necessary to produce 1 mol of H2. Thus, we can write:
2 mol HCl ------------------- 1 mol H2
0.274 mol HCl ------------ y
Solving for y, we have that 0.137 moles of H2 would be obtained.
We can convert the number of moles of H2 to its correspondent mass as it follows (the molar mass of H2 is 2.016 g/mol):
[tex]\begin{gathered} number\text{ of moles = }\frac{mass\text{ of sample \lparen g\rparen}}{molar\text{ mass \lparen g/mol\rparen}}\rightarrow mass\text{ of sample = number of moles}\times molar\text{ mass} \\ \\ mass\text{ of H}_2=0.137mol\times2.016g/mol=0.276g \end{gathered}[/tex]Therefore, 0.276g of H2 could be obtained from the reaction, considering the mass of reactants given.
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.
