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Nata [24]
2 years ago
7

Suppose 650 mL of hydrogen gas are produced through a displacement reaction involving solid iron and sulfuric acid, H2SO4, at ST

P. How many grams of iron (II) sulfate are produced? Density H2(g) is on your periodic table.
Chemistry
1 answer:
Ainat [17]2 years ago
8 0

Answer:

                     Mass  =  4.415 g of FeSO₄

Explanation:

                   The balance chemical equation for given single replacement reaction is as follow;

                                  Fe + H₂SO₄ → FeSO₄ + H₂

Data Given;

                  Volume =  650 mL  =  0.65 L

                  Density at STP =  0.08988 g/L

                  Mass = Density × Volume = 0.08988 g/L × 0.65 L = 0.0584 g

Step 1: <u>Calculate Moles of H₂ as;  </u>

                  Moles  =  Mass / M.Mass

                  Moles  =  0.0584 g / 2.01 g/mol

                  Moles  =  0.0290 mol of H₂

Step 2: <u>Find out moles of FeSO₄ as;</u>

According to balance chemical equation,

            1 mole of H₂ is produced along with  =  1 mole of FeSO₄

So,

   0.0290 moles of H₂ will be produced along with  =  X moles of FeSO₄

Solving for X.

                     X  =  0.0290 × 1 mol / 1 mol

                     X =  0.0290 moles of FeSO₄

Step 3: <u>Calculate mass of FeSO₄ as;</u>

                   Mass  =  Moles × M.Mass

                   Mass  =  0.0290 mol × 151.90 g/mol

                   Mass  =  4.415 g of FeSO₄

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Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 30.0 °C for the following reacti
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Answer : The value of K for this reaction is, 2.6\times 10^{15}

Explanation :

The given chemical reaction is:

CH_3OH(g)+CO(g)\rightarrow HCH_3CO_2(g)

Now we have to calculate value of (\Delta G^o).

\Delta G^o=G_f_{product}-G_f_{reactant}

\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]

where,

\Delta G^o = Gibbs free energy of reaction = ?

n = number of moles

\Delta G^0_{(HCH_3CO_2(g))} = -389.8 kJ/mol

\Delta G^0_{(CH_3OH(g))} = -161.96 kJ/mol

\Delta G^0_{(CO(g))} = -137.2 kJ/mol

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

\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]

\Delta G^o=-89.4kJ/mol

The relation between the equilibrium constant and standard Gibbs, free energy is:

\Delta G^o=-RT\times \ln K

where,

\Delta G^o = standard Gibbs, free energy  = -89.4 kJ/mol = -89400 J/mol

R = gas constant  = 8.314 J/L.atm

T = temperature  = 30.0^oC=273+30.0=303K

K = equilibrium constant = ?

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

-89400J/mol=-(8.314J/L.atm)\times (303K)\times \ln K

K=2.6\times 10^{15}

Thus, the value of K for this reaction is, 2.6\times 10^{15}

4 0
3 years ago
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