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Sloan [31]
2 years ago
6

If 1.2 g of ammonium bicarbonate is treated with .75 grams of sodium chloride, 63.0075 grams sodium bicarbonate will be produced

. In this reaction NaCI is limiting reagent. True or false
Chemistry
1 answer:
Ivan2 years ago
6 0

Answer:

True

Explanation:

Limiting reagent is the reactant that determines the progress of the reaction. It determines how much of the product is formed.

The equation for this reaction is;

NaCl (aq) + NH4HCO3 (aq) → NaHCO3 (aq) + NH4Cl (aq)

From the reaction, 1 mol of NaCl reacts with 1 mol of NH4HCO3 to produce 1 mol of NaHCO3

Converting to masses using; Mass = Number of moles * Molar mass

58.44g of NaCl reacts with 79.056g of NH4HCO3

If we were to  sue the whole 1.2 g of NH4HCO3 we would require xg of NaCl

58.44 = 79.056

x = 1.2

x = 0.887g

The fact that 1.2g of NH4HCO3 requires 0.887g of NaCl (which is less than the available 0.75g) means that NaCl is the limiting reagent.

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How much energy must be removed from a 125 g sample of benzene (molar mass= 78.11 g/mol) at 425.0 K to liquify the sample and lo
riadik2000 [5.3K]

Answer : The energy removed must be, -67.7 kJ

Solution :

The process involved in this problem are :

(1):C_6H_6(g)(425.0K)\rightarrow C_6H_6(g)(353.0K)\\\\(2):C_6H_6(g)(353.0K)\rightarrow C_6H_6(l)(353.0K)\\\\(3):C_6H_6(l)(353.0K)\rightarrow C_6H_6(l)(335.0K)

The expression used will be:

\Delta H=[m\times c_{p,g}\times (T_{final}-T_{initial})]+m\times \Delta H_{vap}+[m\times c_{p,l}\times (T_{final}-T_{initial})]

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\Delta H = heat released by the reaction = ?

m = mass of benzene = 125 g

c_{p,g} = specific heat of gaseous benzene = 1.06J/g^oC

c_{p,l} = specific heat of liquid benzene = 1.73J/g^oC

\Delta H_{vap} = enthalpy change for vaporization = 33.9kJ/mole=33900J/mole=\frac{33900J/mole}{78.11g/mole}J/g=434.0J/g

Molar mass of benzene = 78.11 g/mole

Now put all the given values in the above expression, we get:

\Delta H=[125g\times 1.06J/g.K\times (353.0-(425.0))K]+125g\times -434.0J/g+[125g\times 1.73J/g.K\times (335.0-353.0)K]

\Delta H=-67682.5J=-67.7kJ

Therefore, the energy removed must be, -67.7 kJ

4 0
2 years ago
A 30.0 g sample of a metal was heated in a hot water bath to 80°c. it was then quickly transferred to a coffee-cup calorimeter.
Annette [7]
when the metal  is lost heat and the calorimeter of water is gained the heat 

and when the heat lost = the heat gained so,

(M*C*ΔT)m =  (M*C*ΔT)w

when Mm= mass of the metal = 30 g 

Δ Tm = (80-25) = 55 °C

and Mw = mass of water = 100 g  

Cw is the specific heat of water = 4.181 J/g.°C

ΔTw = (25-20) = 5 °C

so by substitution:

∴ 30* Cm*55 = 100 * 4.181 * 5 

∴Cm (specific heat of metal) = (100*4.181*5)/(30*55) 

∴C of metal = 1.267 J/g.°C
3 0
2 years ago
Read 2 more answers
The ph of a solution that contains 0.818 m acetic acid (ka = 1.76 x 10-5) and 0.172 m sodium acetate is __________.
crimeas [40]
PH is calculated using <span>Handerson- Hasselbalch equation,

                                      pH  =  pKa  +  log [conjugate base] / [acid]

Conjugate Base  =  Acetate (CH</span>₃COO⁻)
Acid                    =  Acetic acid (CH₃COOH)
So,
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We are having conc. of acid and acetate but missing with pKa,
pKa is calculated as,

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                                     pKa  =  4.75
Now,
Putting all values in eq. 1,
                     
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Answer:

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Answer:

Follows are the solution to this question:

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3 0
2 years ago
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