Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, hcl(aq) , as described by the chemical equation mno2(s)+4hcl(aq)⟶mncl2(aq)+2h2o(l)+ cl2(g) how much mno2(s) should be added to excess hcl(aq) to obtain 255 ml cl2(g) at 25 °c and 725 torr ?

0.86 g

Explanation:

Given:  

Pressure = 725 torr

The conversion of P(torr) to P(atm) is shown below:

So,  

Pressure = 725 / 760 atm = 0.9539 atm

Volume = 225 mL = 0.225 L (1 mL = 0.001 L)

Temperature = 25 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (25 + 273.15) K = 298.15 K  

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.9539 atm × 0.255 L = n × 0.0821 L.atm/K.mol × 298.15 K  

⇒n = 0.0099 moles

From the reaction,

1 mole of chlorine is formed from 1 mole of

Thus, moles of = 0.0099 moles

Molar mass of = 86.9368 g/mol

Mass = Moles*Molar mass = 0.0099 moles * 86.9368 g/mol = 0.86 g

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2