Following a chemical reaction that produced 5.06 grams of magnesium chloride, the lab report was prepared to document the results. The expected result was estimated to be 8.85 grams. What are the percent yield and percent
error that are to be included in the lab report?
Percent Yield
Percent Error

1. Percentage yield = 57.2%

2. Percentage error = 74.9%

Explanation:

From the question given above, the following data were obtained:

Actual yield = 5.06 g

Experimental yield = 8.85 g

Percentage yield =?

Percentage error =?

1. Determination of the percentage yield.

Actual yield = 5.06 g

Experimental yield = 8.85 g

Percentage yield =?

Percentage yield = Actual yield /Experimental yield × 100

Percentage yield = 5.06 / 8.85 × 100

Percentage yield = 57.2%

2. Determination of the Percentage error.

Actual yield = 5.06 g

Experimental yield = 8.85 g

Percentage error =?

Percentage error = |Experimental – Actual| / Actual yield × 100

Percentage error = |8.85 – 5.06| / 5.06 × 100

Percentage error = 3.79 / 5.06 × 100

Percentage error = 74.9%

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

d.) the average kinetic energy of the gas particles is greater in container b because its particles move faster.