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tia_tia [17]
3 years ago
6

A laboratory technician combined sodium hydroxide with excess iron(II) nitrate. A reaction took place according to this chemical

equation: 2NaOH + Fe(NO3)2 → NaNO3 + Fe(OH)2.
The reaction produced 3.70 grams of iron(II) hydroxide.

Assuming the reaction came to completion, what was the initial mass of sodium hydroxide? Use the periodic table.

A. 1.6 g
B. 2.0 g
C. 3.3 g
D. 4.0 g
Chemistry
1 answer:
Whitepunk [10]3 years ago
8 0

 The initial  mass  of sodium hydroxide  is  3.3 g (answer C)

 <u><em>calculation</em></u>

Step 1 : find the  moles of iron (ii) hydroxide (  Fe(OH)₂

moles =  mass÷  molar mass

from periodic table the  molar mass of Fe(OH)₂  = 56 + [16 +1]2  = 90 g/mol

moles  is therefore = 3.70 g÷ 90 g/mol = 0.041 moles

Step 2:  use the mole ratio to  calculate the moles of  sodium hydroxide (NaOH)

   from given equation  NaOH : Fe(OH)₂    is 2 :1

therefore the moles of NaOH = 0.041 x 2 = 0.082 moles

Step 3: find  mass of NaOH

mass = moles x molar mass

from the periodic table the  molar mass of NaOH = 23 +16 +1  = 40 g/mol

mass  = 0.082  moles x 40 g/mol = 3.3 g ( answer C)


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If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcohol evaporate? If some liquid remains, how much will there be? The vapor pressure of ethyl alcohol at 25 °C is 59 mm Hg, and the density of the liquid at this temperature is 0.785g/cm^3 .

will all the alcohol evaporate? or none at all?

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Yes, all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore  be zero.

Explanation:

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The volume of alcohol which is placed in a small laboratory = 1.0 L

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Converting 59 mmHg to atm ; since 1 atm = 760 mmHg;

Then, we have:

= \frac{59}{760}atm

= 0.078 atm

Temperature = 25 ° C

= ( 25 + 273 K)

= 298 K.

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The volume of laboratory = l × b × h

= 3.0 m × 2.0 m × 2.5 m

= 15 m³

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Using ideal gas equation to determine the moles of ethanol in vapor phase; we have:

PV = nRT

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n = \frac{0.078 * 15000}{0.082*290}

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Moles of ethanol in 1.0 L bottle can be calculated as follows:

Since  numbers of moles = \frac{mass}{molar mass}

and mass = density × vollume

Then; we can say ;

number of moles = \frac{density*volume }{molar mass of ethanol}

number of moles =\frac{0.785g/cm^3*1000cm^3}{46.07g/mol}

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