Question

A student obtained the following data in this experiment. Fill in the blanks in the data and make the indicated calculations

Mass of gelatin capsule 0.1284 g Ambient temperature, t 22°C
Mass of capsule plus alloy sample 0.3432 g Ambient temperature, T ____________ K
Mass of alloy sample, m ____0.2148________ g Barometric pressure 745 mm Hg Mass of empty beaker 144.5 g Vapor pressure of H2O at t __________ mm Hg
Mass of beaker plus displaced water 368.9 g Pressure of dry H2, ____745________ mm Hg Pressure of dry H2 ____________ atm
Mass of displaced water _____224.4_______ g Volume of displaced water (density = 1.00 g/mL) ____________ mL Volume, V, of H2 = Volume of displaced water ____________ mL; ____________ liters

Find the number of moles of H2 evolved, (V in liters, P in atm, T in K, R = 0.0821 literatm/ mole K). ____________ moles H2
Find , the number of moles of H2 per gram of sample.

Answer 1

To find the number of moles of H2, we use the Ideal Gas Law equation PV = nRT and substitute the given values. The number of moles of H2 per gram of sample can be calculated by dividing the moles of H2 by the mass of the sample.

Step-by-step explanation:

To find the number of moles of H2 evolved, we can use the Ideal Gas Law equation: PV = nRT. In this case, we are given the volume of H2 (2.00 L), the temperature (35.0°C which we need to convert to Kelvin, so 35 + 273 = 308 K), the pressure (7.41 x 107 N/m² which we need to convert to atm, so 7.41 x 107 / 1.013 x 105 = 731 atm), and the gas constant R (0.0821 literatm/ mole K).

Substituting these values into the equation, we can solve for n:

n = (PV) / (RT) = (731 atm x 2.00 L) / (0.0821 literatm/ mole K x 308 K) = 38.17 moles H2

The number of moles of H2 per gram of sample can be calculated by dividing the moles of H2 by the mass of the sample:

moles H2 per gram of sample = (38.17 mol H2) / (0.2148 g sample) = 177.74 mol/g sample

Answer 2

Ambient temperature K = 295 K

Pressure of dry H2 (atm) = 0.980 atm

Volume of displaced water (mL) = 224.4 mL

Volume of displaced water in liters = 0.2244 L

Number of mole of H2 = 0.00908 mol

Moles of H2 per gram sample = 0.042 mol H2/g

Step-by-step explanation:

Given data:

Mass of gelatin capsule = 0.1284 g; Ambient temperature (T) = 22°C

Mass of capsule plus alloy sample = 0.3432g; Ambient temperature K = ?

Ambient temperature K = 22+273

= 295 K

Mass of alloy sample = 0.2148; Barometric pressure 745 mm Hg

Mass of beaker plus displaced water =368.9 g

Pressure of dry H2 = 745 mm Hg

Pressure of dry H2 in atm = ?

Pressure of dry H2 (atm) = 745 mmHg *1 atm/760 mmHg

= 0.980 atm

Mass of displaced water = 224.4 g

density = 1.00 g/mL)

Volume of displaced water = ?

Density = mass/volume

Volume = mass/density

=224.4 g/1. g/mL

= 224.4 mL

Therefore, volume of displaced water = 224.4 mL

Volume of displaced water in liters = 224.4*10^-3 L

= 0.2244 L

Calculating the number of mole of H2 using ideal gas law, we have;

PV = nRT

n = PV/TT

where;

n = number of mole of H2

P = pressure of H2 = 0.980 atm

V = volume of H2 = 0.2244 L

R = gas constant = 0.0821 literatm/ mole K

T = temperature = 295 K

Substituting into the formula, we have;

n = PV/RT

= 0.980* 0.2244/ 0.0821*295

= 0.219912/24.2195

= 0.00908 mol

Therefore, number of mole of H2 = 0.00908 mol

Calculating the number of moles of H2 per gram sample using the formula;

Moles of H2 per gram sample = Moles of H2/Mass of sample

= 0.00908 /0.2148

= 0.042 mol H2/g