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dem82 [27]
3 years ago
9

When ch4(g) reacts with h2o(g) to form h2(g) and co(g), 206 kj of energy are absorbed for each mole of ch4(g) that reacts. write

a balanced thermochemical equation for the reaction with an energy term in kj as part of the equation. note that the answer box for the energy term is case sensitive. use the smallest integer coefficients possible and put the energy term in the last box on the appropriate side of the equation. if a box is not needed, leave it blank. + + + +?
Chemistry
1 answer:
jeka57 [31]3 years ago
7 0
1) Chemical reaction: CH₄ + H₂O → 3H₂ + CO ΔH = + 206 kJ/mol.
2) Chemical reaction: CO + H₂O → CO₂ + H₂, ΔH = + 2,8 kJ/mol.
3) Chemical reaction: CH₄ + 2O₂ → CO₂ + 2H₂O ΔH = -802 kJ/mol.
4) Chemical reaction: 2H₂ + O₂ → 2H₂O ΔH = 2·(-242 kJ7mol) = - 484 kJ/mol.
Endothermic reaction (ΔH>o) and exothermic reaction (ΔH<span><0).</span>
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if 334.6 g of phosphoric acid is reacted with excess potassium hydroxide. the final mass K3PO4 produced is found to be 248g. wha
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Answer:

                     %age Yield  =  34.21 %

Explanation:

                   The balance chemical equation for the decomposition of KClO₃ is as follow;

                            3 KOH + H₃PO₄ → K₃PO₄ + 3 H₂O

Step 1: Calculate moles of H₃PO₄ as;

Moles = Mass / M/Mass

Moles = 334.6 g / 97.99 g/mol

Moles = 3.414 moles

Step 2: Find moles of K₃PO₄ as;

According to equation,

                 1 moles of H₃PO₄ produces  =  1 moles of K₃PO₄

So,

              3.414 moles of H₃PO₄ will produce  =  X moles of K₃PO₄

Solving for X,

                      X = 1 mol × 3.414 mol / 1 mol

                      X = 3.414 mol of K₃PO₄

Step 3: Calculate Theoretical yield of K₃PO₄ as,

Mass = Moles × M.Mass

Mass = 3.414 mol × 212.26 g/mol

Mass = 724.79 g of K₃PO₄

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%age Yield  =  Actual Yield / Theoretical Yield × 100

%age Yield  =  248 g / 724.79 × 100

%age Yield  =  34.21 %

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Answer:

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Explanation:

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In this case, since a dilution process implies that the moles of the solute remain the same before and after the addition of diluting water, we can write:

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M_2=\frac{M_2V_2}{V_1} =\frac{45.45mL*0.0404M}{550.00mL}\\\\M_2=3.34x10^{-3}M

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