Starting with 0.657 g of lead(II) nitrate, a student collects 0.925 g of precipitate. If the calculated mass of precipitate is 0.914 g, what is the percent yield

101.2%

Explanation:

Given:

Theoretical yield of the precipitate = 0.914 g

Actual yield of the precipitate = 0.925 g

Now, the percent yield is given as a ratio of actual yield by theoretical yield expressed as a percentage.

Framing in equation form, we have:

Now, plug in 0.925 g for actual yield, 0.914 g for theoretical yield and solve for % yield. This gives,

Therefore, the percent yield is 101.2%.

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.