Question

Какой объем H2 выделится при взаимодействии 20г Mg и 60г HCl ?? ПОМОГИТЕ ПОЖАЛУЙСТА СРОЧНО РЕШИТЬ

Answer 1

Mg+2hcl=mgcl2+h2
находим кол-во вещества магния и соляной кислоты
20/24=0,84 моль
60/36=1,65 моль
n(mg)/n(hcl)=1/2 следовательно mg в избытке, считаем по молям hcl
2n(hcl)=n(h2)=0,825 моль
1 моль любого газа занимает 22,4 литра следовательно 
0,825*22,4=18,48 литров

Answer 2

Calculation of Hydrogen Gas Volume

To calculate the volume of hydrogen gas (H2) released during the reaction between 20g of magnesium (Mg) and 60g of hydrochloric acid (HCl), we need to use the concept of stoichiometry and the molar masses of the substances involved.

1. Determine the molar masses: - The molar mass of magnesium (Mg) is 24.31 g/mol. - The molar mass of hydrochloric acid (HCl) is 36.46 g/mol.

2. Convert the given masses to moles: - Moles of Mg = mass of Mg / molar mass of Mg = 20g / 24.31 g/mol - Moles of HCl = mass of HCl / molar mass of HCl = 60g / 36.46 g/mol

3. Write the balanced chemical equation for the reaction: - The balanced equation for the reaction between Mg and HCl is: Mg + 2HCl → MgCl2 + H2.

4. Determine the stoichiometric ratio: - From the balanced equation, we can see that 1 mole of Mg reacts with 2 moles of HCl to produce 1 mole of H2.

5. Calculate the moles of H2 produced: - Moles of H2 = moles of Mg × (moles of H2 / moles of Mg) = moles of Mg × 1/1 = moles of Mg.

6. Convert moles of H2 to volume: - The volume of 1 mole of any gas at standard temperature and pressure (STP) is 22.4 liters. - Therefore, the volume of H2 gas produced is equal to the moles of H2.

Let's perform the calculations:

Molar mass of Mg = 24.31 g/mol Molar mass of HCl = 36.46 g/mol

Moles of Mg = 20g / 24.31 g/mol = 0.822 mol Moles of HCl = 60g / 36.46 g/mol = 1.644 mol

Moles of H2 = Moles of Mg = 0.822 mol

Volume of H2 gas produced = Moles of H2 = 0.822 mol

Therefore, the volume of hydrogen gas (H2) released during the reaction is 0.822 liters.

Please note that the above calculation assumes the reaction takes place at standard temperature and pressure (#### Calculation of Hydrogen Gas Volume

To determine the volume of hydrogen gas (H2) that will be produced when 20 grams of magnesium (Mg) reacts with 60 grams of hydrochloric acid (HCl), we need to use the stoichiometry of the balanced chemical equation for the reaction.

The balanced chemical equation for the reaction between magnesium and hydrochloric acid is as follows:

Mg + 2HCl → MgCl2 + H2

From the balanced equation, we can see that 1 mole of magnesium reacts with 2 moles of hydrochloric acid to produce 1 mole of hydrogen gas.

To calculate the moles of magnesium and hydrochloric acid, we need to divide the given masses by their respective molar masses.

The molar mass of magnesium (Mg) is approximately 24.31 g/mol. The molar mass of hydrochloric acid (HCl) is approximately 36.46 g/mol.

Calculation Steps:

1. Calculate the moles of magnesium (Mg): - Moles of Mg = Mass of Mg / Molar mass of Mg - Moles of Mg = 20 g / 24.31 g/mol

2. Calculate the moles of hydrochloric acid (HCl): - Moles of HCl = Mass of HCl / Molar mass of HCl - Moles of HCl = 60 g / 36.46 g/mol

3. Determine the limiting reactant: - The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product that can be formed. - To determine the limiting reactant, compare the moles of Mg and HCl. The reactant with the smaller number of moles is the limiting reactant.

4. Calculate the moles of hydrogen gas (H2) produced: - Moles of H2 = Moles of limiting reactant (either Mg or HCl) * (1 mole of H2 / 1 mole of Mg or HCl)

5. Convert the moles of hydrogen gas to volume using the ideal gas law: - Volume of H2 = Moles of H2 * Molar volume of gas at STP (standard temperature and pressure) - Molar volume of gas at STP is approximately 22.4 L/mol.

Calculation:

1. Moles of Mg: - Moles of Mg = 20 g / 24.31 g/mol = 0.822 mol

2. Moles of HCl: - Moles of HCl = 60 g / 36.46 g/mol = 1.645 mol

3. Determining the limiting reactant: - The moles of Mg (0.822 mol) are less than the moles of HCl (1.645 mol), so Mg is the limiting reactant.

4. Moles of H2 produced: - Moles of H2 = Moles of limiting reactant (Mg) * (1 mole of H2 / 1 mole of Mg) - Moles of H2 = 0.822 mol * (1 mole of H2 / 1 mole of Mg) = 0.822 mol

5. Volume of H2 produced: - Volume of H2 = Moles of H2 * Molar volume of gas at STP - Volume of H2 = 0.822 mol * 22.4 L/mol = 18.4 L

Therefore, approximately 18.4 liters of hydrogen gas will be produced when 20 grams of magnesium reacts with 60 grams of hydrochloric acid.

Please note that this calculation assumes the reaction takes place at standard temperature and pressure (

Calculating the volume of H2 gas produced

The question asks us to calculate the volume of hydrogen gas (H2) produced from the reaction of 20 grams of magnesium (Mg) with 60 grams of hydrochloric acid (HCl).

The balanced chemical equation for this reaction is:

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

Using the mole ratios from the balanced equation:

- 20g Mg = 0.8333 moles Mg - 60g HCl = 0.6667 moles HCl - The mole ratio is 1 mole Mg to 2 moles HCl - So there is excess HCl of 0.3333 moles - By the mole ratio, this excess HCl can produce 0.3333 moles of H2 gas

Using the gas law PV=nRT to calculate the volume:

- n = 0.3333 moles H2 gas - R = 0.0821 L·atm/mol·K (ideal gas constant) - T = 298 K (room temperature) - P = 1 atm (standard pressure)

Plugging into the gas law gives: V = nRT/P V = (0.3333 mol) * (0.0821 L·atm/mol·K) * (298 K) / (1 atm) V = 10 L of H2 gas