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MrMuchimi
2 years ago
12

If you start with 0.020 g of Mg, how many moles of H2 will you make if the reaction is complete?

Chemistry
1 answer:
irina [24]2 years ago
6 0

If one starts with 0.020 g of Mg, 0.0008 moles of H2 would be made if the reaction is complete.

Going by the balanced equation of reaction in the image, 1 mole of Mg will produce 1 mole of H2 in a complete reaction.

If 0.020 g of Mg is started with:

mole of Mg = mass/molar mass

                        = 0.020/24.3

                            = 0.0008 moles

Since the mole of Mg to H2 is 1:1, thus, 0.0008 moles of H2 will also be made from the reaction.

More on stoichiometry can be found here: brainly.com/question/9743981

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or a particular isomer of C 8 H 18 , the combustion reaction produces 5104.1 kJ of heat per mole of C 8 H 18 ( g ) consumed, und
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Answer:

The standard enthalpy of formation of this isomer of C_{8}H_{18} is -220.1 kJ/mol.

Explanation:

The given chemical reaction is as follows.

C_{8}H_{18}(g)+ \frac{25}{2}O_{2}(g) \rightarrow 8CO_{2}(g)+9H_{2}O(g)

\Delta H^{o}_{rxn}= -5104.1kJ/mol

The expression for the entropy change for the reaction is as follows.

\Delta H^{o}_{rxn}=[8\Delta H^{o}_{f}(CO_{2}) +9\Delta H^{o}_{f}(H_{2}O)]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}\Delta H^{o}_{f}(O_{2})]

\Delta H^{o}_{f}(H_{2}O)= -241.8kJ/mol

\Delta H^{o}_{f}(CO_{2})= -393.5kJ/mol

\Delta H^{o}_{f}(O_{2})= 0kJ/mol

Substitute the all values in the entropy change expression.

-5104.1kJ/mol=[8(-393.5)+9(-241.8)kJ/mol]-[\Delta H^{o}_{f}(C_{8}H_{18})+ \frac{25}{2}(0)kJ/mol]

-5104.1kJ/mol=-5324.2kJ/mol -\Delta H^{o}_{f}(C_{8}H_{18})

\Delta H^{o}_{f}(C_{8}H_{18}) =-5324.2kJ/mol +5104.1kJ/mol

=-220.1kJ/mol

Therefore, The standard enthalpy of formation of this isomer of C_{8}H_{18} is -220.1 kJ/mol.

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