What volume of hydrogen at STP is produced from the reaction of 50.0 g of Mg and the equivalent of 75g of HCl?
Well, let's see... If we're talking about the reaction between magnesium (Mg) and hydrochloric acid (HCl), we can use the balanced chemical equation:
Mg + 2HCl -> MgCl2 + H2
According to the equation, one mole of magnesium reacts with 2 moles of hydrochloric acid to produce one mole of hydrogen gas. So, first we need to determine the number of moles of Mg and HCl.
The molar mass of Mg is approximately 24.3 g/mol, and the molar mass of HCl is approximately 36.5 g/mol.
Therefore, with 50.0 g of Mg, we can find the number of moles:
50.0 g Mg * (1 mol Mg / 24.3 g Mg) = approximately 2.06 moles of Mg
With 75 g of HCl, we can find the number of moles of HCl:
75 g HCl * (1 mol HCl / 36.5 g HCl) = approximately 2.05 moles of HCl
Since the reaction is 1:2 between Mg and HCl, the limiting reactant is HCl. That means 2.05 moles of HCl will react to produce 1 mole of hydrogen gas.
At STP (standard temperature and pressure), 1 mole of any gas occupies approximately 22.4 liters.
Therefore, the volume of hydrogen gas produced would be:
1 mole H2 * 22.4 L/mol = 22.4 liters of hydrogen gas.
So, approximately 22.4 liters of hydrogen gas would be produced from this reaction. But remember, this answer is rather serious, not clown-worthy. I apologize if it wasn't quite what you expected!
To determine the volume of hydrogen gas produced from the reaction of magnesium (Mg) and hydrochloric acid (HCl) at STP (standard temperature and pressure), we can follow these steps:
Step 1: Write and balance the chemical equation for the reaction:
Mg + 2HCl -> MgCl2 + H2
From the balanced equation, we can see that 1 mole of Mg reacts with 2 moles of HCl to produce 1 mole of H2.
Step 2: Calculate the number of moles of Mg:
Given mass of Mg = 50.0 g
Molar mass of Mg = 24.31 g/mol
Number of moles of Mg = mass of Mg / molar mass of Mg
= 50.0 g / 24.31 g/mol
≈ 2.06 moles
Step 3: Calculate the number of moles of HCl:
Given mass of HCl = 75 g
Molar mass of HCl = 36.46 g/mol
Number of moles of HCl = mass of HCl / molar mass of HCl
= 75 g / 36.46 g/mol
≈ 2.06 moles
Note: The stoichiometric ratio of Mg to HCl is 1:2, which means that for every 1 mole of Mg reacting, 2 moles of HCl are required.
Step 4: Determine the limiting reagent:
To find the limiting reagent, we compare the number of moles of Mg and HCl and identify which one is present in a lesser amount. The reactant that is in a lesser stoichiometric amount will determine the amount of product formed.
In this case, both Mg and HCl have the same number of moles, i.e., 2.06 moles. Therefore, there is a 1:1 ratio between the reactants, and none of them is in excess.
Step 5: Calculate the number of moles of H2 produced:
Since the stoichiometric ratio between Mg and H2 is also 1:1, the number of moles of H2 produced will be equal to the number of moles of Mg, which is 2.06 moles.
Step 6: Calculate the volume of H2 at STP:
1 mole of any ideal gas at STP occupies 22.4 L of volume.
Therefore, the volume of hydrogen gas produced at STP is:
Volume of H2 = number of moles of H2 × molar volume at STP
= 2.06 moles × 22.4 L/mol
≈ 46.5 L
So, the volume of hydrogen gas produced from the reaction of 50.0 g of Mg and the equivalent of 75 g of HCl at STP is approximately 46.5 L.
To find the volume of hydrogen gas produced at STP (Standard Temperature and Pressure), you need to use the concept of the ideal gas law. The ideal gas law equation is given as:
PV = nRT
Where:
P = pressure (which is the STP pressure, 1 atmosphere)
V = volume
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (which is the STP temperature, 273 K)
To solve this problem, we need to follow these steps:
Step 1: Determine the number of moles of magnesium (Mg)
To find the number of moles of Mg, you need to divide the given mass (50.0 g) by the molar mass of magnesium (24.31 g/mol). The molar mass of Mg is found on the periodic table.
Number of moles of Mg = 50.0 g / 24.31 g/mol
Step 2: Determine the number of moles of hydrogen chloride (HCl)
Similar to step 1, find the number of moles of HCl by dividing 75 g by the molar mass of HCl (which is the sum of the atomic masses of hydrogen and chlorine).
The molar mass of hydrogen is approximately 1 g/mol, and the molar mass of chlorine is approximately 35.45 g/mol.
Number of moles of HCl = 75 g / (1 g/mol + 35.45 g/mol)
Step 3: Determine the limiting reactant
To determine the limiting reactant, compare the moles of Mg with the moles of HCl. The reactant that has the lesser number of moles is the limiting reactant, as it will be completely consumed in the reaction.
Step 4: Calculate the number of moles of hydrogen gas produced
The balanced chemical equation for the reaction between Mg and HCl is:
Mg + 2HCl → MgCl2 + H2
Based on the balanced equation, for every 1 mole of Mg, 1 mole of hydrogen gas is produced.
Step 5: Calculate the volume of hydrogen gas at STP
Since 1 mole of an ideal gas occupies 22.4 L at STP, you can use this conversion factor to find the volume of hydrogen gas produced.
Volume of hydrogen gas (at STP) = Moles of hydrogen gas × 22.4 L/mol
By following these steps, you can find the volume of hydrogen gas produced at STP from the given reactants.
This is a limiting reagent (LR) problem. You know that when quantities are given for both reactanats.
Mg + 2HCl --> H2 + MgCl2
mols Mg = g/atomic mass = 50/24.3 = estimated 2
mols HCl = g/molar mass = 75/36.5 = estimated 2
1 mol Mg, according to the equation, reacts with 2 mols HCl so 2 mols Mg will require 4 mols HCl. You don't have that much HCl; therefore, HCl is the LR. So estimated 2 mols HCl will produce estimated 1 mol H2 gas. 1 mol H2 gas @ STP occupies 22.4 L. You should go through all of the calculations (not estimated) to obtain a more accurate number.