Mg + 2HCl ----> MgCl2 + H2
What volume of hydrogen at STP is produced from the reaction of 50.0g of Mg and the equivalent if 75g of HCl?
1. What is the limiting reagent?
2. What is the volume of hydrogen produced?
3. How much is in excess?
But what does "volume=22.4*1.04 dm^3" mean???
I'm new to chemistry so don't know the lingo that well. Would it be 23.3 excess?
molesMg=50/26=1.92
molesHCl=75/36=2.08
clearly you do not have enougth HCl to consume all the Mg.
Limiting Reagent: HCl
volume H2: moles H2=1/2 moles HCL=1.04
volume=22.4*1.04 dm^3 that is the volume in decimeters^3 (some still call that a liter)
excess excess. I assume that means excess Mg, as it was not all used up.
You used: moles HCl=2.08, or then moles Mg 1.04*26 or 27 grams, so you have 50-27 g of Mg left over, or in excess then of 23 grams.
Awesome thank you I wanted to make sure I understand you meant. So it looks like we got the same answer of excess= 23g Mg
Thanks!
To answer these questions, we need to follow a step-by-step process. Let's go through each question one by one:
1. What is the limiting reagent?
To determine the limiting reagent, we will compare the amount of each reactant used in the reaction to their stoichiometric ratio. The stoichiometric ratio is the balanced equation coefficients that indicate the mole ratio of the reactants.
The balanced equation is:
Mg + 2HCl -> MgCl2 + H2
First, we need to convert the given masses of Mg and HCl into moles by using their molar masses. The molar mass of Mg is approximately 24.31 g/mol, and the molar mass of HCl is approximately 36.46 g/mol.
For Mg:
moles of Mg = mass of Mg / molar mass of Mg
moles of Mg = 50.0 g / 24.31 g/mol
For HCl:
moles of HCl = mass of HCl / molar mass of HCl
moles of HCl = 75 g / 36.46 g/mol
Now we can compare the mole ratio of Mg to HCl. From the balanced equation, we see that the ratio is 1:2. This means that 1 mole of Mg reacts with 2 moles of HCl. Let's calculate the actual ratio:
moles of Mg / ratio of Mg to HCl = moles of HCl / ratio of Mg to HCl
moles of Mg / 1 = moles of HCl / 2
Simplifying the equation, we get:
2 * moles of Mg = moles of HCl
If the number of moles of Mg is equal to or greater than half the number of moles of HCl, then Mg is the limiting reagent. Otherwise, HCl is the limiting reagent.
2. What is the volume of hydrogen produced?
To calculate the volume of hydrogen, we need to use the ideal gas law at STP (Standard Temperature and Pressure). At STP, 1 mole of any gas occupies 22.4 liters.
First, we need to determine the number of moles of hydrogen produced. From the balanced equation, we see that the stoichiometric ratio is 1:1 between Mg and H2. Therefore, the number of moles of H2 is equal to the number of moles of Mg.
Next, we can use the mole-to-volume conversion:
volume of H2 in liters = moles of H2 * 22.4 L/mol
3. How much is in excess?
To determine the amount in excess, we need to compare the number of moles of the non-limiting reagent (the reactant that is not completely consumed) with the stoichiometric ratio.
If the number of moles of the non-limiting reagent is greater than what is required by the stoichiometry, it is in excess. To convert the excess moles to grams, we use the molar mass of the corresponding substance.
Once we have the grams of excess reactant, we can find the volume using the ideal gas law at STP (1 mole = 22.4 liters) for gases.
I hope this explanation helps you in solving the problem step by step. Let me know if you have any further questions.