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Alisiya [41]
2 years ago
6

Hợp chất X gồm 2 nguyên tố A và O.Trong đó A cs hóa trị IV.A chiếm 50% về khối lượng.

Chemistry
1 answer:
astraxan [27]2 years ago
8 0

Answer:

A.lap cong thic hoa hoc cua A:va H

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miv72 [106K]

The percent yield of the calcium hydroxide is 84.5%.

<h3>What is stoichiometry?</h3>

Stoichiometry enables us to obtain the mass of a substance form the equation of the reaction.

The equation of the reaction is;

CaCO3 + 2HCl -----> CaCl2 + CO2 + H2O

Number of moles of X = 40.0 grams/100 g/mol = 0.4 moles

Number of moles of HCl = 850/1000 * 1 M = 0.85 moles

If 1 mole of CaCO3 reacts with 2 moles of HCl

0.4 moles of  CaCO3 reacts with  0.4 moles  * 2 moles/1 mole

= 0.8 moles of HCl

Hence X is the limiting reactant.

The reaction is 1:1 then the amount of CO2  produced is 0.4 moles

Mass of CO2 = 0.4 CO2 * 44 g/mol = 17.6 g

2) The reaction equation is; 2NaOH + CaCO3 --->  Ca(OH)2 + Na2CO3

Number of moles of X = 25.0 grams/100 g/mol =  0.25 moles

Number of moles of NaOH= 40/1000 L * 2 M = 0.08 moles

If 1 mole of X reacts with 2 moles of NaOH

0.25 moles  reacts with   0.25 moles   * 2 moles /1 mole

= 0.5 moles

NaOH is the limiting reactant

2 moles of NaOH produces 1 mole of CO2

0.08 moles of NaOH produces 0.08 moles * 1 mole/2 moles

= 0.04 moles of CO2

Theoretical yield of CO2 =  0.04 moles of CO2 * 74 g/mol = 2.96  g

Percent yield = 2.5 g/ 2.96  g * 100

= 84.5%

Learn ore about percent yield:brainly.com/question/17042787

#SPJ1

5 0
1 year ago
Michelle is trying to find the average atomic mass of a sample of an unknown
GREYUIT [131]

The average atomic mass of her sample is 114.54 amu

Let the 1st isotope be A

Let the 2nd isotope be B

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

  • Abundance of isotope A (A%) = 59.34%
  • Mass of isotope A = 113.6459 amu
  • Mass of isotope B = 115.8488 amu
  • Abundance of isotope B (B%) = 100 – 59.34 = 40.66%
  • Average atomic mass =?

The average atomic mass of the sample can be obtained as follow:

Average \: atomic \: mass \:  =  \frac{mass \: of \: A \times A\%}{100}  + \frac{mass \: of \: B \times B\%}{100}  \\  \\ Average \: atomic \: mass \:  =  \frac{113.6459\times 59.34}{100} + \frac{115.8488\times 40.66}{100} \\  \\ Average \: atomic \: mass \:  = 114.54 \: amu  \\  \\

Thus, the average atomic mass of the sample is 114.54 amu

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