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Genrish500 [490]
3 years ago
15

The combustion of methane gas (ch4) forms co2(g)+ h2o(ℓ). calculate the heat produced by burning 2 mol of the methane gas. use t

hese ∆h0 f data to help: ch4(g)= -74.9 kj/mol co2(g)= -393.5kj/mol h2o(ℓ)= -285.8kj/mol. answer in units of kj.
Chemistry
1 answer:
Nostrana [21]3 years ago
5 0
<span>Hess's Law is as follows. Note: Google Hess's law for the actual formula (i.e. with the Greek characters). ΔH(reaction) = Sum of ΔH(formation of products) - Sum of ΔH(formation of reactants) Now determine you're formula/reaction. In this case it's as follows: CH4 + 2 O2 -----> CO2 + 2 H2O The ΔH(formation) can be found in standardized tables, either online, in books, or provided to you. The ΔH you need for this is as follows in kJ/mol: CH4 (g) -75 O2 (g) 0 CO2 (g) -394 H2O (l) -286 When you plug the info into the formula you get the following: ΔH(reaction) = [(2*-238) + -394)] - [-75 + (2*0)] ΔH(reaction) = -891 kJ/mol Now you're looking for the amount of kJ for 1.65 mol. Your standard formula is as follows: E = ΔH(reaction) * mol where E is energy measured in kJ. So when you plug in your relevant values you get: E = -891 * 1.65 E = -1470.15 kJ</span>
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What is the molarity of a solution that contains 0.0345 mol nh4cl in exactly 400 ml of solution?
alexdok [17]
The molarity of a solution is the number of moles of a substance per liter of the solution.
400 mL becomes 0.400 liters
You have 0.0345 moles of ammonium chloride NH4Cl, so take the number of moles over the number of liters to find the answer.
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4 0
3 years ago
2 C6H14 + 19 O2 --&gt; 12 CO2 + 14 H2O
Rashid [163]

Answer:

4.06 mol H₂O

Explanation:

  • 2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O

First we <em>convert the given masses of reactants into moles</em>, using <em>their respective molar masses</em>:

  • 250 g O₂ ÷ 32 g/mol = 7.81 mol O₂
  • 50 g C₆H₁₄ ÷ 86 g/mol = 0.58 mol C₆H₁₄

Now we <u>calculate how many O₂ moles would react completely with 0.58 C₆H₁₄ moles</u>, using the <em>stoichiometric coefficients of the reaction</em>:

  • 0.58 mol C₆H₁₄ * \frac{19molO_2}{2molC_6H_{14}} = 5.51 mol O₂

As there are more O₂ moles than required (7.81 vs 5.51), O₂ is the reactant in excess. That means that <em>C₆H₁₄ is the limiting reactant</em>.

Now we can <u>calculate how much water can be formed</u>, using <em>the number of moles of the limiting reactant</em>:

  • 0.58 mol C₆H₁₄ * \frac{14molH_2O}{2molC_6H_{14}} = 4.06 mol H₂O
5 0
3 years ago
Apply scientific knowledge and understanding to determine how many grams of carbon dioxide are produced from the combustion of 2
lidiya [134]

Answer:

749 grams CO₂

Explanation:

To find the amount of carbon dioxide produced, you need to (1) convert grams C₃H₈ to moles C₃H₈ (via molar mass from periodic table), then (2) convert moles C₃H₈ to moles CO₂ (via mole-to-mole ratio via reaction coefficients), and then (3) convert moles CO₂ to grams CO₂ (via molar mass from periodic table). It is important to arrange the ratios/conversions in a way that allows for the cancellation of units. The desired unit should be in the numerator. The final answer should have 3 significant figures because the given value (250. grams) has 3 sig figs.

Molar Mass (C₃H₈): 3(12.01 g/mol) + 8(1.008 g/mol)

Molar Mass (C₃H₈): 44.094 g/mol

1 C₃H₈ + 5 O₂ ---> 3 CO₂ + 4 H₂O

Molar Mass (CO₂): 12.01 g/mol + 2(16.00 g/mol)

Molar Mass (CO₂): 44.01 g/mol

250. g C₃H₈         1 mole C₃H₈           3 moles CO₂              44.01 g
------------------  x  ----------------------  x  ----------------------  x  --------------------  =
                               44.094 g              1 mole C₃H₈            1 mole CO₂

= 749 grams CO₂

6 0
2 years ago
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