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Pavlova-9 [17]
3 years ago
5

A student is interested in determining the number of water molecules in a hydrate called manganese(II) chloride. (MnCl2). She kn

ow that the number of water molecules in a hydrate is usually a small whole number from 1 to 5. She also knows that the chemical formula of a hydrate can be determined by analyzing the percent water in the substance. The student prepared a reference table of possible % water values for comparison to her experimental results.
Hydrate MnCl2⋅H2O MnCl2⋅2H2O MnCl2⋅3H2O MnCl2⋅4H2O MnCl2⋅5H2O
MM of anhydrous salt 125.84 125.84 125.84 125.84 125.84
MM of nH2O 18.02 36.04 54.06 72.08 90.10
MM of hydrate 143.86 262.88 179.90 197.92 215.94
Percent water 12.52 22.26 30.05 36.42 41.72

With this in mind, she weighs out six 1-g samples of her unknown salt, puts each sample in a different test tube, and then uses a Bunsen burner to evaporate the water out of each sample. Next, she measures the mass of the solid left in each test tube.

Trial Mass of Sample Mass of Solid Left in the tube Mass of Evaporated water Percent Water in the Hydrate
1 1.00 g 0.69 0.31 31%
2 1.00 g 0.63 0.37 37%
3 1.00 g 0.67 0.33 33%
4 1.00 g 0.65 0.35 35%
5 1.00 g 0.67 0.33 33%
6 1.00 g 0.64 0.36 36%

The unknown hydrate was manganese(II) chloride is a trihydrate, MnCl2⋅3H2O. The mass of the water evaporated out of the hydrate ranged from 0.31 g to 0.37 g. which means the percent water in sample is about 34%. Since the amount of water in MnCl2⋅3H2O should be 30% of the total mass, the manganese(II) chloride must be a trihydrate.

a. Review the data and argument provided. The student's claim is INCORRECT.
b. Examine the data tables and think about what you know about reactions, mole ratios, chemical composition and empirical formulas.
c. Briefly discuss the student's argument and provide empirical evidence and/or reasons to demonstrate why her claim is inaccurate. Be sure to include what you think the student's experimental error might have been.
d. Clearly state the claim you would make based on the evidence and a sufficient rationale to support it. Be sure to present your ideas in a clear and logical order, using a variety of words and well constructed sentences.
Chemistry
1 answer:
Phantasy [73]3 years ago
8 0

Answer:

Following are the responses to the given choices.

Explanation:

For point a:

Its idea of the test is to calculate water through unknown hydrate that has been evaporated on the burn.

MnCl_2.nH2O\longrightarrow MnCl_2 + nH_2O

This mole ratio is abnormal to the sample amount of water vaporized. Each student is thus able to examine only 1 g of sample. The amount of water is then linked to a standard table that she has already planned.

Its empirical definition of MnCl_2.nH_2O to n variation of 1 to 5 was distinct and the water percentage was determined by mass.  If higher the amount of hydrate, the higher the amount of water in salt.

Its results vary from 31\% \ to \ 37\% of water content. Its water slides for three moisturize molecules are 30\% and the water slides for four hydrate molecules are 36.4\%. Her argument which four hydrate molecules could therefore be eliminated could be inaccurate.

For point b:

The error could be incomplete fire, which could not have vaporized all hydrate compounds.Its student may not have evaporated all swimming pools in the salt, because the estimate of their moisture content could be on the bottom.

Its average moisture content is roughly 34\%. Consequently, it should logically search for water content close to over 34\%. In addition to the average value, thus, the salt is composed of 4 moisturize molecules of approximately 36\%.

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60.0 g O2 and 50.0 g S are reacted according to the equation 2 S + 3 O2 → 2 SO3 . Which reactant is in excess and by how many gr
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excess reactant sulpher and mass left=10g

Explanation:

first write ha balance chemical equation:

2 S + 3 O_2 \rightarrow 2 SO_3

mole=\frac{given \, mass}{molar\,mass}

Given mass of sulfur=50g

molar mass=32 g/mol

mole of sulfur=1.5625 mol

given mass of O_2=60g

molar mass=32 g/mol

mole of oxygen=1.875mol

from the above balanced equation it is clearly that;

2 mole of sulfur reacts completely with 3 mole of Oxygen

1 mole of sulfur reacts completely with 1.5 mole of Oxygen

1.5625 mole of sulfur will react with 1.5×1.5625 (2.34375 mol)mole of oxygen

but we have only 1.875 mole of oxgygen hence sulpfur will be the excess reactant and oxygen will be thelimiting reactant

therefore,

3 mole of Oxygen reacts completely with 2 mole of sulfur

1 mole of Oxygen reacts completely with 2/3 mole of sulfur

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1.875 mole of Oxygen will react completely with (2/3)×1.875(1.25mol) mole of sulfur

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