If 95.0 g Zn is added to an excess of H2SO4, how many liters of hydrogen gas will be produced at STP?
Write and balance the equation. Then, calculate the
liters of hydrogen produced.

Please help

32.7 L of H₂

Explanation:

We'll begin by calculating the number of mole in 95 g of Zn. This can be obtained as follow:

Mass of Zn = 95 g

Molar mass of Zn = 65 g/mol

Mole of Zn =?

Mole = mass /Molar mass

Mole of Zn = 95/65

Mole of Zn = 1.46 mole

Next, we shall determine the number of mole of H₂ produced by the reaction of 95 g (i.e 1.46 mole) of Zn. This can be obtained as illustrated below:

Zn + H₂SO₄ —> ZnSO₄ + H₂

From the balanced equation above,

1 mole of Zn reacted to produce 1 mole of H₂.

Therefore, 1.46 mole of Zn will also react to produce 1.46 mole of H₂.

Finally, we shall determine the volume of H₂ obtained at STP. This can be obtained as follow:

1 mole of H₂ = 22.4 L at STP.

Therefore,

1.46 mole of H₂ = 1.46 × 22.4

1.46 mole of H₂ = 32.7 L at STP

Thus, 32.7 L of H₂ were obtained from the reaction.

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.