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Vedmedyk [2.9K]
3 years ago
12

In the reaction h2so4 + na2co3 → na2so4 + co2 + h2o 14.5 g of water is collected. what is the molarity of sulfuric acid if 1.78

l were used in the reaction? 1. 146 m 2. 0.453 m 3. 1.43 m 4. 0.697 m 5. 1.78 m6. 0.0046 m
Chemistry
1 answer:
Yuliya22 [10]3 years ago
8 0
H2SO4 + Na2CO3  → Na2SO4 + CO2 + H2O

The molarity  of  sulfuric acid  if  1.78  L  were used  in the   above reaction is
 0.453 M  (answer 2)

      Calculation
find the  moles of water  produced  = mass/molar  mass

=   14.5 g /18 g/mol  = 0.806  moles

by use of of mole ratio  between  H2So4  to H2O   which is 1:1  the    moles    of   H2SO4   is also =  0.806   moles


Molarity  of H2SO4 is therefore =    moles/volume  in   liters

=  0.806   mol/  1.78 L  =  0.453 M (answer 2)
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1. find the molar mass (amu) of each element and add them to get the whole molar mass.

2. divide the 1 element molar mass with the whole molar mass

3. multiple by 100 and that gives you the % composition.

<h2><u><em>56-57: NaCl</em></u></h2>

1. Na(22.99amu) + Cl (35.453amu)=58.443

2(Na):   \frac{22.99}{58.443} = .393

2(Cl): \frac{35.453}{58.443}= .607

3(Na): .393 * 100=39.3%

3(Cl): .607 * 100= 60.7%

<h2><u>58-60 </u>K_{2} CO_{3}<u /></h2>

1. K: (39.098)(2)=78.196

_ C: (12.011)(1)= 12.011

_O: (15.99)(3) = 47.997

78.196+12.011+47.997= 138.204

2:K: \frac{78.196}{138.204}= .566 <u>Step </u>3: (.566)(100)= 56.6%

2: C: \frac{12.011}{138.204}= .087 <u>Step 3</u>: (.087)(100)= 8.7%

2: O: \frac{47.997}{138.204}= .347 <u>Step 3</u>: (.347)(100) = 34.7%

<h2>61-62 Fe_{3} O_{4}</h2>

1. Fe (55.845)(3)= 167.535

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2: Fe: \frac{167.535}{231.531}= .724 Step 3: (.724)(100)= 72.4%

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<h2>63-65 C_{3}H_{5}(OH)_{3}</h2>

1.

C(12.011*3)=36.033

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