In the lab, a student calculated and average P x V value of 48.0 mmHgʻL using the syringe-pressure gauge apparatus at a constant temperature used in this experiment. Calculate the number of moles
of air in this system if the P x V value was measured at a constant 18.0°C.
(Hint: The Ideal Gas Law has the variables of P, V, n&T; R = 62.37mmHg•L/mol•K)

0.0026 mole

Explanation:

The following data were obtained from the question:

Pressure (P) × volume (V) = P × V = 48 mmHg•L

Temperature (T) = 18.0°C.

Gas constant (R) = 62.37 mmHg•L/mol•K

Number of mole (n) =.?

Next, we shall convert 18 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

T(°C) = 18 °C

T(K) = 18 + 273

T(K) = 291 K

Finally, we shall determine the number of mole of air in the system as shown below:

Pressure (P) × volume (V) = P × V = 48 mmHg•L

Temperature (T) = 291 K.

Gas constant (R) = 62.37 mmHg•L/mol•K

Number of mole (n) =.?

PV = nRT

48 = n × 62.37 × 291

48 = n × 18149.67

Divide both side by 18149.67

n = 48 / 18149.67

n = 0.0026 mole

Therefore, the number of mole of air in the system is 0.0026 mole

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

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