Question

V(H2)=4.48л m(Mg)-? Mg+2HCl=MgCl2+H2

Answer 1

Ответ:

Объяснение:

 Х г                                                       4,48 л

 Mg     +     2HCl     =     MgCl2     +     H2

n=1 моль                                            n=1 моль

М=24 г/моль                                   Vm=22,4 л/моль

m=24 г                                             V = 22,4 л

Х г   Mg    -    4,48 л  Н2

24 г  Mg   -  22,4 л  Н2

m(Mg) = 24 * 4,48 / 22,4 = 4,8 г

Answer 2

The balanced chemical equation for the reaction between magnesium (Mg) and hydrochloric acid (HCl) is:

Mg + 2HCl → MgCl2 + H2

According to the equation, 1 mole of magnesium reacts with 2 moles of hydrochloric acid to produce 1 mole of magnesium chloride and 1 mole of hydrogen gas.

To find the volume of hydrogen gas evolved when 4.48 liters of magnesium reacts, we need to calculate the moles of magnesium using the ideal gas law equation:

PV = nRT

Where: P = Pressure (in atm) V = Volume (in liters) n = Number of moles R = Ideal gas constant (0.0821 L·atm/(mol·K)) T = Temperature (in Kelvin)

Assuming the reaction takes place at standard temperature and pressure (STP), which is 273.15 K (0°C) and 1 atm, respectively, we can rearrange the ideal gas law equation to solve for the number of moles:

n = PV / RT

Using the values for STP, we can calculate the number of moles of magnesium:

n(Mg) = (1 atm) * (4.48 L) / (0.0821 L·atm/(mol·K) * 273.15 K) = 0.187 mol

According to the balanced equation, 1 mole of magnesium produces 1 mole of hydrogen gas. Therefore, the number of moles of hydrogen gas formed will also be 0.187 mol.

Now, to find the volume of hydrogen gas, we can use the ideal gas law equation again. Assuming the conditions are still STP, we have:

V(H2) = n(H2) * RT / P = (0.187 mol) * (0.0821 L·atm/(mol·K)) * (273.15 K) / (1 atm) = 4.88 L

Therefore, the volume of hydrogen gas evolved when 4.48 liters of magnesium reacts is approximately 4.88 liters.