How many grams of AlCl3 are needed to completely react with 2.25 of NaOH?

Explanation:

hope the picture above help u understand I did it in step so it would be easier to understand:)

1- 114.54 g.

2- 78.32 g.

Explanation:

Q1:

The balanced equation of the reaction is:

F₂ + 2NaCl → Cl₂ + 2NaF.

It is clear that 1.0 mole of F₂ reacts with 2.0 moles of NaCl to produce 1.0 mole of Cl₂ and 2.0 moles of NaF.We can calculate the no. of moles of F₂ (15.0 liters at 280.0 K and 1.50 atm) using the ideal gas law:

PV = nRT,

where, P is the pressure of the gas (P = 1.0 atm),

V is the volume of the gas (V = 15.0 L),

n is the no. of moles of the gas (n = ??? mole),

R is the general gas constant (R = 0.082 L.atm/mol.K),

T is the temperature of the gas (T = 280.0 K).

∴ n = PV/RT = (1.5 atm)(15.0 L) / (0.082 L/atm/mol.K)(280.0 K) = 0.98 mol.

We can get the no. of moles of NaCl  reacted with 0.98 mol of F₂.

Using cross multiplication:

1.0 mole of F₂ reacts with → 2.0 moles of NaCl, from the stichiometry.

0.98 mole of F₂ reacts with → ??? moles of NaCl.

∴ The no, of moles of reacted NaCl = (2.0 mole)(0.98 mole)/(1.0 mole) = 1.96 mol.

Now, we can get the mass of NaCl that reacted with F₂ at 280.0 K and 1.50 atm:

∴ The mass of NaCl = n x molar mass of NaCl = (1.96 mol)(58.44 g/mol) = 114.54 g.

Q2:

STP conditions means that P = 1.0 atm and T = 273.0 K.We can calculate the no. of moles of F₂ (15.0 liters at 273.0 K and 1.0 atm) using the ideal gas law: PV = nRT,

∴ n = PV/RT = (1.0 atm)(15.0 L) / (0.082 L/atm/mol.K)(273.0 K) = 0.67 mol.

We can get the no. of moles of NaCl  reacted with 0.67 mol of F₂.

Using cross multiplication:

1.0 mole of F₂ reacts with → 2.0 moles of NaCl, from the stichiometry.

0.67 mole of F₂ reacts with → ??? moles of NaCl.

∴ The no, of moles of reacted NaCl = (2.0 mole)(0.67 mole)/(1.0 mole) = 1.34 mol.

Now, we can get the mass of NaCl that reacted with F₂ at 280.0 K and 1.50 atm:

∴ The mass of NaCl = n x molar mass of NaCl = (1.34 mol)(58.44 g/mol) = 78.32 g.