Please answer quick. In an experiment, a student collects 107 mL of H2 over water at a pressure of 104.8 kPa and a temperature of 30°C. How many grams of H2 are present? A.
0.024 g
B.
4.00 g
C.
0.04 g
D.
0 049 g
And explain

Mass = 0.008 g

Explanation:

Given data:

Volume of H₂ = 107 mL

Pressure = 104.8 KPa

Temperature = 30°C

Mass of H₂ present = ?

Solution:

Formula:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

Now we will convert the units of temperature, pressure and volume.

Temperature  = 30+273 = 303 K

Volume of H₂ = 107 mL (107/1000= 0.107 L)

Pressure = 104.8 KPa (104.8/101 = 1.04 atm)

by putting values,

1.04 atm ×0.107 L = n× 0.0821 atm.L/ mol.K  × 303 K

0.11 atm.L  = n×24.87 atm.L/ mol

n =  0.11 atm.L / 24.87 atm.L/ mol

n = 0.004

Mass of H₂:

Mass = number of moles × molar mass

Mass =   0.004    mol × 2 g/mol

Mass = 0.008 g

the dependant variable is spilled water

the independent is the heat