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kumpel [21]
3 years ago
9

A chemist requires 0.752 mol na2co3 for a reaction. how many grams does this correspond to?

Chemistry
1 answer:
Sliva [168]3 years ago
8 0
<span>79.70 grams First, calculate the molar mass of Na2CO3 by looking up the molar mass of the elements used in it. Sodium = 22.989769 Carbon = 12.0107 Oxygen = 15.999 Multiply each molar mass by the number of atoms used, and sum the results to get the molar mass 2 * 22.989769 + 12.0107 + 3 * 15.999 =105.9872 Finally, multiply the molar mass by the number of moles you need. 0.752 * 105.9872 = 79.7024 grams</span>
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Carbon, hydrogen and ethane each burn exothermically in an excess of air. AHⓇ =-393.7 kJ mol. C(s) + O2(g) → CO2(g) H2(g) + % O2
Salsk061 [2.6K]

<u>Answer:</u> The \Delta H^o_{rxn} for the reaction is 51.8 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The chemical equation for the reaction of carbon and water follows:

2C(s)+2H_2(g)\rightarrow C_2H_4(g) \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) C(s)+O_2(g)\rightarrow CO_2(g)    \Delta H_1=-393.7kJ    ( × 2)

(2) H_2+\frac{1}{2}O_2(g)\rightarrow H_2O(l)    \Delta H_2=-285.9kJ     ( × 2)

(3) 2C_2H_4(s)+2O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)    \Delta H_3=-1411kJ

The expression for enthalpy of the reaction follows:

\Delta H^o_{rxn}=[2\times \Delta H_1]+[2\times \Delta H_2]+[1\times (-\Delta H_3)]

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(2\times (-393.7))+(2\times (-285.9))+(1\times -(-1411))]=51.8kJ

Hence, the \Delta H^o_{rxn} for the reaction is 51.8 kJ.

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