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tigry1 [53]
4 years ago
7

For the reaction shown, 38.25 grams of silicon is combined with 14.33 grams of nitrogen gas. How many grams of silicon nitride c

an be formed if nitrogen is the limiting reactant? 3Si + 2N2 yields Si3N4 17.94 g 21.23 g 35.88 g 63.69 g
Chemistry
1 answer:
mr_godi [17]4 years ago
6 0
From the periodic table:
molar mass of silicon = 28 grams
molar mass of nitrogen = 14 grams

From the balanced chemical equation, 
3 x 28 = 84 grams of silicon react with 2 x 2 x 14 = 56 grams of nitrogen to produce 3(28) + 4(14) = 140 grams of silicon nitride.

Since it is mentioned that nitrogen is the limiting reagent, therefore, the reaction will stop once the amount of reacting nitrogen is consumed.
Now we know that 56 grams of nitrogen produce 140 grams of silicon nitride. Thus, we can use cross multiplication to calculate the amount of silicon nitride produced when using 14.33 grams of nitrogen as follows:

amount of produced silicon nitride = (14.33 x 140) / 56 = 35.8 grams


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Answer:

A

Explanation:

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Use standard enthalpies of formation to calculate Δ H ∘ rxn for each reaction. MISSED THIS? Read Section 7.9; Watch KCV 7.9, IWE
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Answer:

Standard Heat of Reaction 1 = -136.2 kJ/mol

Standard Heat of Reaction 2 = -41.166 kJ/mol

Standard Heat of Reaction 3 = -136.07 kJ/mol

Standard Heat of Reaction 4 = 279.448kJ/mol

Explanation:

C₂H₄ (g) + H₂ (g) → C₂H₆ (g)

CO (g) + H₂O (g) → H₂ (g) + CO₂ (g)

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The required standard heat of formation for each of the reactants and product above, as obtained from literature is listed below.

C₂H₄ (g), 52.5 kJ/mol

H₂ (g), 0 kJ/mol

C₂H₆ (g), -83.7 kJ/mol

CO (g), -110.525 kJ/mol

H₂O (g), -241.818 kJ/mol

H₂ (g), 0 kJ/mol

CO₂ (g), -393.509 kJ/mol

NO₂ (g), 33.2 kJ/mol

H₂O (l), -285.8 kJ/mol

HNO₃ (aq), -206.28 kJ/mol

NO (g), 90.29 kJ/mol

Cr₂O₃ (s), -1128.4 kJ/mol

CO (g), -110.525 kJ/mol

Cr (s), 0 kJ/mol

CO₂ (g), -393.509 kJ/mol

Note that

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

C₂H₄ (g) + H₂ (g) → C₂H₆ (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (1×-83.7) = -83.7 kJ/mol

ΔH∘(reactants) = (1×52.5) + (1×0) = 52.5 kJ/mol

ΔH∘(rxn) = -83.7 - 52.5 = -136.2 kJ/mol

CO (g) + H₂O (g) → H₂ (g) + CO₂ (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (1×0) + (1×-393.509) = -393.509 kJ/mol

ΔH∘(reactants) = (1×-110.525) + (1×-241.818) = -352.343 kJ/mol

ΔH∘(rxn) = -393.509 - (-352.343) = -41.166 kJ/mol

3NO₂ (g) + H₂O (l) → 2HNO₃ (aq) + NO (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (2×-206.28) + (1×90.29) = -322.27 kJ/mol

ΔH∘(reactants) = (3×33.2) + (1×-285.8) = -186.2 kJ/mol

ΔH∘(rxn) = -322.27 - (-186.2) = -136.07 kJ/mol

Cr₂O₃ (s) + 3CO (g) → 2Cr (s) + 3CO₂ (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (2×0) + (3×-393.509) = -1,180.527 kJ/mol

ΔH∘(reactants) = (1×-1128.4) + (3×-110.525) = -1,459.975 kJ/mol

ΔH∘(rxn) = -1,180.527 - (-1,459.975) = 279.448 kJ/mol

Hope this Helps!!!

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