Avolume of 129 ml of hydrogen is collected over water. the water level in the collecting vessel is the

Explanation:

As it is given that water level is same as outside which means that theoretically, P = 756.0 torr.

So, using ideal gas equation we will calculate the number of moles as follows.

                  PV = nRT

or,           n = [tex]\frac{PV}{RT}[/tex]

                 = [tex]\frac{\frac{756}{760}atm \times 0.129 L}{0.0821 Latm/mol K \times 298 K}[/tex]

                  = 0.0052 mol

Also,  No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]

               0.0052 mol = [tex]\frac{mass}{2 g/mol}[/tex]

                  mass = 0.0104 g

As some of the water over which the hydrogen gas has been collected is present in the form of water vapor. Therefore, at [tex]25^{o}C[/tex]

                [tex]P_{\text{water vapor}}[/tex] = 24 mm Hg

                                = [tex]\frac{24}{760}[/tex] atm

                                = 0.03158 atm

Now,   P = [tex]\frac{756}{760} - 0.03158[/tex]

              = 0.963 atm

Hence,   n = [tex]\frac{0.963 atm \times 0.129 L}{0.0821 L atm/mol K \times 298 K}[/tex]

                 = 0.0056 mol

So, mass of [tex]H_{2}[/tex] = 0.0056 mol × 2

                         = 0.01013 g (actual yield)

Therefore, calculate the percentage yield as follows.

      Percent yield = [tex]\frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100[/tex]

                              = [tex]\frac{0.01013 g}{0.0104 g} \times 100[/tex]            

                              = 97.49%

Thus, we can conclude that the percent yield of hydrogen for the given reaction is 97.49%.

The molar mass of metal is 52.4 g/mol

Explanation:

To calculate the moles of hydrogen gas collected, we use the equation given by ideal gas which follows:

[tex]PV=nRT[/tex]

where,

P = pressure of the hydrogen gas = Total pressure - vapor pressure of water = (756.0 - 23.8 ) torr = 732.2 torr  

V = Volume of the gas = 201 mL = 0.201 L    (Conversion factor:  1 L = 1000 mL)

T = Temperature of the gas = [tex]25^oC=[25+273]K=298K[/tex]

R = Gas constant = [tex]62.364\text{ L.mmHg }mol^{-1}K^{-1}[/tex]

n = number of moles of hydrogen gas = ?

Putting values in above equation, we get:

[tex]732.2torr\times 0.210L=n\times 62.364\text{ L. torr }mol^{-1}K^{-1}\times 298K\\\\n=\frac{732.2\times 0.210}{62.364\times 298}=0.0083mol[/tex]

The given chemical equation follows:

[tex]M(s)+H_ 2SO_4(aq.)\rightarrow MSO_4(aq.)+H_2(g)[/tex]

By Stoichiometry of the reaction:

1 mole of hydrogen gas is formed by 1 mole of metal

So, 0.0083 moles of hydrogen gas will be formed by = [tex]\frac{1}{1}\times 0.0083=0.0083mol[/tex] of metal

To calculate the molar mass of metal from given number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Moles of metal = 0.0083 moles

Given mass of metal = 0.435 g

Putting values in above equation, we get:

[tex]0.0083mol=\frac{0.435g}{\text{Molar mass of metal}}\\\\\text{Molar mass of metal}=\frac{0.435g}{0.0083mol}=52.4g/mol[/tex]

Hence, the molar mass of metal is 52.4 g/mol

55.0 g/mol

Step-by-step explanation:

Step 1. Partial pressure of hydrogen

You are collecting the gas over water, so

[tex]p_{\text{atm}} = p_{\text{H}_{2}} + p_{\text{H}_{2}\text{O}}[/tex]

[tex]p_{\text{H}_{2}} = p_{\text{atm}} - p_{\text{H}_{2}\text{O}}[/tex]

[tex]p_{\text{atm}} = \text{756.0 Torr}[/tex]

At 25 °C,[tex]p_{\text{H}_{2}\text{O}} = \text{23.8 Torr}[/tex]

[tex]p_{\text{H}_{2}} = \text{756.0 Torr} - \text{23.8 Torr} = \text{732.2 Torr}[/tex]

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Step 2. Moles of H₂

We can use the Ideal Gas Law.

pV = nRT                                       Divide both sides by RT

n = (pV)/(RT)

p = 732.2 Torr                               Convert to atmospheres

p = 732.2/760  

p = 0.9634 atm

V = 291 mL                                     Convert to litres

V = 0.291 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 25 °C                                        Convert to kelvins

T = (25 + 273.15 ) K = 298.15 K    

n = (0.9632 × 0.291)/(0.082 06 × 298.15)      

n = 0.2804/24.47

n = 0.011 46 mol

===============

Step 3. Moles of metal

The partial chemical equation is

M + H₂SO₄ ⟶ H₂ + …

The molar ratio of M:H₂ is 1 mol M:1 mol H₂.

Moles of M = 0.011 46× 1/1

Moles of M = 0.011 46 mol M

===============

Step 4. Atomic mass of M

Atomic mass = mass of M/moles of M

Atomic mass = 0.630/0.011 46

Atomic mass = 55.0 g/mol

Explanation:

Several gaseous properties comprise the temperature, volume, pressure, and the number of moles of gases. These properties can be determined from each other through their relationships utilized by the ideal gas law.

Answer and Explanation:

Given:

mass of MM, m=0.502gm=0.502g

volume of hydrogen gas, V=217mL=0.217LV=217mL=0.217L

temperature, TT = 25ooC + 273.15 = 298.15K298.15K

(Using data from http://www.wiredchemist.com/chemistry/data/vapor-pressure)

vapor pressure of water at 25ooC, PwPw = 23.8torr×1atm760torr=0.0313atm23.8torr×1atm760torr=0.0313atm

total pressure, PP = 756torr×1atm760torr=0.9947atm756torr×1atm760torr=0.9947atm

universal gas constant, RR = 0.08205L⋅atmmol⋅K0.08205L⋅atmmol⋅K

The balanced chemical reaction:

MM(s) + H2SO4H2SO4(aq) ⇋⇋ MSO4MSO4(aq) + H2H2(g)

Assuming only water and hydrogen contributes to the total pressure, the partial pressure of hydrogen, PH2PH2, can be calculated from the following equation,

P=PH2+PwP=PH2+Pw

0.9947atm=PH2+0.0313atm0.9947atm=PH2+0.0313atm

PH2=0.9634atmPH2=0.9634atm

Solving for moles of H2H2, nH2nH2, using the ideal gas law,

PH2V=nH2RTPH2V=nH2RT

0.9634atm×

Use the PV = nRt formulaYou find n (you have everything else)then find with molar mass using n