Question

What volume of hydrogen gas (at STP) will be liberated by the reaction of excess Zn on 50 mL dilute H2so4 of specific gravity = 2 (g/ml) and w/w percentage 49% ?

Answer 1

The volume of hydrogen liberated is 6.4 L.

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Answer 2

Answer:

$\huge\mathbf\pink{QuesTion}$

What volume of hydrogen gas (at STP) will be liberated by the reaction of excess Zn on 50 mL dilute H2so4 of specific gravity = 2 (g/ml) and w/w percentage 49% ?

$\huge\mathit\blue{Answer}$

The volume of hydrogen liberated is 6.4L.

Step 1. Calculate the mass of H2SO4

Mass of H2SO4=50mL solution×1.3g solution

1mL solution×40 g H2SO4

100g solution =26.0 g H2SO4

Step 2. Calculate the moles of H2SO4

Moles of H2SO4 = 26.0g H2SO4× 1mol

H2SO498.08g H2SO4 =0.265 mol H2SO4

Step 3. Write the balanced equation for the reaction

Zn + H2SO4→H2+ZnSO4

Step 4. Calculate the moles of H2

Moles of H2=0.265 mol H

2SO4×1 mol H

21mol H

2SO4 =0.265 mol H2

Step 5. Calculate the volume of hydrogen

We can use the Ideal Gas Law to calculate the volume:

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

apV= nR Ta

$a∣∣$

−−−−−−−−−−−−−−−

We can rearrange the Ideal Gas Law to get

V= nRTp

Since NTP is 20 °C and 1 atm, we have

n=0.265 mol

R=0.082 06 L⋅ atm ⋅K -1 mol-1T

=(20 + 273.15) K = 293.15 Kp=1 atm

V=0.265mol×0.082 06 L⋅atm

⋅

K-1 mol

-1×293.15K

1atm=6.4 L

hope it's helpful for you .

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