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BlackZzzverrR [31]
3 years ago
5

if you react 100g of ammonium chloride with excess calcium oxide, what Is the theoretical yeild (in grams) of ammonia? when you

complete the reaction, your actual yeild of ammonia is 8.12g. what is your percent yeild for the production of ammonia using this process​
Chemistry
1 answer:
Maurinko [17]3 years ago
4 0

Answer:

31.78 grams

25.55%

Explanation:

The balanced reaction for ammonium chloride with calcium oxide will be:

2NH4Cl + Ca(OH)2 ---> CaCl2 + 2NH3 + 2H2O

The molecular weight for ammonium chloride(NH4Cl ) is 53.49g/mol, while the molecular weight for ammonium(NH3) is 17g/mol. The number of theoretical yield of ammonia from 100g of ammonium chloride will be:

100g / (53.49g/mol) * 2/2  * 17g/mol= 31.78 grams

If the actual yield is 8.12g, the percent yield will be: 8.12g/31.78g * 100% =25.55%

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Americans eat too much salt (NaCl). A healthy adult should eat no more than 6 g of salt in one day. Use this information to answ
Rufina [12.5K]

Answer:

a. 58.5 g/mol

b. 0.1 mol

Explanation:

a.

The molar mass of Na is 23.0 g/mol. The molar mass of Cl is 35.5 g/mol. The molar mass of NaCl is:

M(Na) + M(Cl) = 23.0 g/mol + 35.5 g/mol = 58.5 g/mol

b. A healthy adult should eat no more than 6 g of salt in one day. The moles corresponding to 6 g of NaCl are:

6 g × (1 mol/58.5 g) = 0.1 mol

7 0
3 years ago
A mixture of methane and carbon dioxide gases contains methane at a partial pressure of 431 mm Hg and carbon dioxide at a
KatRina [158]

Answer:

XCH₄ = 0.461

XCO₂ = 0.539

Explanation:

Step 1: Given data

  • Partial pressure of methane (pCH₄): 431 mmHg
  • Partial pressure of carbon dioxide (pCO₂): 504 mmHg

Step 2: Calculate the total pressure in the container

We will sum both partial pressures.

P = pCH₄ + pCO₂

P = 431 mmHg + 504 mmHg = 935 mmHg

Step 3: Calculate the mole fraction of each gas

We will use the following expression.

Xi = pi / P

XCH₄ = pCH₄/P = 431 mmHg/935 mmHg = 0.461

XCO₂ = pCO₂/P = 504 mmHg/935 mmHg = 0.539

3 0
3 years ago
The conversion of cyclopropane to propene occurs with a first-order rate constant of . How long will it take for the concentrati
Gennadij [26K]

This is an incomplete question, here is a complete question.

The conversion of cyclopropane to propene occurs with a first-order rate constant of 2.42 × 10⁻² hr⁻¹. How long will it take for the concentration of cyclopropane to decrease from an initial concentration 0.080 mol/L to 0.053 mol/L?

Answer : The time taken will be, 17.0 hr

Explanation :

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 2.42\times 10^{-2}\text{ hr}^{-1}

t = time passed by the sample  = ?

a = initial concentration of the reactant  = 0.080 M

a - x = concentration left = 0.053 M

Now put all the given values in above equation, we get

t=\frac{2.303}{2.42\times 10^{-2}}\log\frac{0.080}{0.053}

t=17.0\text{ hr}

Therefore, the time taken will be, 17.0 hr

5 0
3 years ago
Consider the following reaction between calcium oxide and carbon dioxide: CaO(s)+CO2(g)→CaCO3(s) A chemist allows 14.4 g of CaO
sweet-ann [11.9K]

Answer:

Theoretical yield =26.03 g

Percent yield = 87%

Limiting reactant = CaO

Explanation:

Given data:

Mass of CaO = 14.4 g

Mass of CO₂ = 13.8 g

Actual yield of CaCO₃ = 22.6 g

Theoretical yield = ?

Percent yield = ?

Limiting reactant = ?

Solution:

Chemical equation:

CaO + CO₂   → CaCO₃

Number of moles of CaO:

Number of moles  = Mass /molar mass

Number of moles = 14.4 g / 56.1 g/mol

Number of moles  = 0.26 mol

Number of moles of CO₂:

Number of moles = Mass /molar mass

Number of moles = 13.8 g / 44 g/mol

Number of moles = 0.31 mol

Now we will compare the moles of CO₂ and CaO with CaCO₃ .

                  CO₂         :                CaCO₃  

                  1               :                 1

                 0.31           :              0.31

                CaO           :               CaCO₃  

                 1                :                 1

                 0.26         :              0.26

The number of moles of  CaCO₃ produced by CaO are less it will be limiting reactant.

Mass of CaCO₃: Theoretical yield

Mass of CaCO₃ = moles × molar mass

Mass of CaCO₃ =0.26 mol × 100.1 g/mol

Mass of CaCO₃ =  26.03 g

Percent yield:

Percent yield = actual yield / theoretical yield × 100

Percent yield = 22.6 g/ 26.03 g × 100

Percent yield = 0.87× 100

Percent yield = 87%

Limiting reactant:

The number of moles of  CaCO₃ produced by CaO are less it will be limiting reactant.

7 0
3 years ago
Please help me with this. (: I'd appreciate it.
Soloha48 [4]

Answer:

The answer to your question is:  ΔH = -283 kJ/mol, first option

Explanation:

Reaction

                CO  +  O₂     ⇒     CO₂

ΔH = ∑H products - ∑H products

ΔH = -393.5 - (-110.5 + 0)

ΔH = -393.5 + 110.5

ΔH = -283 kJ/mol

4 0
3 years ago
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