How many liters of chlorine gas at 25°C and 0.950 atm can be produced by the reaction of 12.0 g of MnO2 with excess HCl(aq) according to the following chemical equation? MnO2(s) + 4HCl(aq) → MnCl2(aq) + 2H2O(l) + Cl2(g)

3.55 L.

Explanation:

We'll begin by calculating the number of mole in 12 g of MnO2. This can be obtained as follow:

Molar mass of MnO2 = 55 + (16×2)

= 55 + 32

= 87 g/mol

Mass of MnO2 = 12 g

Mole of MnO2 =...?

Mole = mass /Molar mass

Mole of MnO2 = 12 / 87

Mole of MnO2 = 0.138 mole

Next, we shall determine the number of mole Cl2 produced from the reaction. This is illustrated below:

The balanced equation for the reaction is given below:

MnO2(s) + 4HCl(aq) → MnCl2(aq) + 2H2O(l) + Cl2(g)

From the balanced equation above,

1 mole of MnO2 reacted to produce 1 mole of Cl2.

Therefore, 0.138 mole of MnO2 will also produce 0.138 mole of Cl2.

Finally, we shall determine the volume of Cl2 gas obtained from the reaction. This can be obtained as shown below:

Temperature (T) = 25 °C = 25 °C + 273 = 298 K

Pressure (P) = 0.950 atm

Number of mole (n) = 0.138 mole

Gas constant (R) = 0.0821 atm.L/Kmol

Volume (V) =.?

PV = nRT

0.950 × V = 0.138 × 0.0821 × 298

Divide both side by 0.950

V = (0.138 × 0.0821 × 298) / 0.950

V = 3.55 L

Therefore, 3.55 L of chlorine gas were obtained from reaction.

if you mean hydrogen and oxygen then water is the product

you didnt put any of the following