I cannot figure out how to work this stoichiometry problem with the given equation: Mg+2HCl-->MgCl2+H2
What volume of hydrogen is produced from the reaction of 50.0g of Mg and the equivalent of 75g of HCl
What do you not understand about it?
It is a limiting reagent problem? Either Mg or HCl will not be completely consumed in the reaction.
Yes. It is a limiting reactant problem. I cannot figure out how you figure out the volume of hydrogen that is produced.
Two ways to do it. The one I like, which takes more time but is easier for me to explain, works this way.
Change g Mg and and g HCl to mols.
mols = g/molar mass.
Convert mols Mg and mols HCl, in two separate steps, to mols H2 produced. Use the coefficients in the balanced equation as usual for this part of the stoichiometry.
The numbers will not be the same. The smaller one (it's always the smaller one) is the one you take. That one, whichever it is; i.e., HCl or Mg, is the limiting reagent and the OTHER one will not be consumed entirely (that is it is the one in excess). See if you can take it from here. Post your work if you get stuck and I can help you through it.
To solve this problem, follow these steps:
1. Convert the given mass of Mg (50.0g) to moles. Use the molar mass of Mg to do this calculation. The molar mass of Mg is 24.31 g/mol. Therefore, moles of Mg = 50.0g / 24.31 g/mol = 2.06 mol of Mg.
2. Convert the given mass of HCl (75g) to moles. Use the molar mass of HCl to do this calculation. The molar mass of HCl is 36.46 g/mol (1 H atom + 1 Cl atom). Therefore, moles of HCl = 75g / 36.46 g/mol = 2.06 mol of HCl.
3. Look at the balanced chemical equation: Mg + 2HCl → MgCl2 + H2. The coefficients tell you the stoichiometric ratio between reactants and products.
4. Use the stoichiometric ratio to determine the limiting reactant. The ratio of Mg to H2 is 1:1, and the ratio of HCl to H2 is 2:1. Comparing the moles of Mg and HCl, we see that both are the same (2.06 mol).
5. Since the stoichiometric ratio of Mg to H2 is 1:1, the moles of H2 produced will be the same as the moles of Mg.
6. Convert the moles of H2 to volume using the ideal gas law equation: PV = nRT. Assume the ideal gas constant R is 0.0821 L·atm/mol·K, and the temperature is constant at room temperature, which is typically around 298K.
P = pressure (assume 1 atm for simplicity)
V = volume of H2 (what we're trying to find)
n = moles of H2 (same as moles of Mg)
R = ideal gas constant
T = temperature in Kelvin
Rearrange the equation to solve for V: V = nRT/P. Plug in the values:
V = 2.06 mol * 0.0821 L·atm/mol·K * 298K / 1 atm.
This simplifies to V = 50.6 L.
Therefore, the volume of hydrogen gas produced from the reaction of 50.0g of Mg and the equivalent of 75g of HCl is 50.6 liters.