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Alex787 [66]
3 years ago
13

This is the chemical formula for nickel tetracarbonyl (a powerfully poisonous liquid used in nickel refining):nico4a chemical en

gineer has determined by measurements that there are 7.0 moles of carbon in a sample of nickel tetracarbonyl. how many moles of oxygen are in the sample?round your answer to 2 significant digits.
Chemistry
1 answer:
lesya [120]3 years ago
8 0

Answer : There are 7.0 moles of oxygen in the formula.

Explanation :

The chemical formula of nickel tetracarbonyl is Ni(CO)₄.

We can see that there are 4 carbon atoms and 4 oxygen atoms which together form 4 carbonyl (CO) groups.

Therefore the mole ratio of C to O is 1 : 1.

Let us use this as a conversion factor to find moles of oxygen.

We have 7.0 moles of carbon in the formula.

The moles of O can be calculated as,

7 mol C \times \frac{1 mol O}{1 mol C} = 7.0 mol O

There are 7.0 moles of oxygen in the formula.

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Given these reactions, X ( s ) + 1 2 O 2 ( g ) ⟶ XO ( s ) Δ H = − 668.5 k J / m o l XCO 3 ( s ) ⟶ XO ( s ) + CO 2 ( g ) Δ H = +
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<u>Answer:</u> The \Delta H^o_{rxn} for the reaction is -1052.8 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 given chemical reaction follows:

X(s)+\frac{1}{2}O_2(g)+CO_2(g)\rightarrow XCO_3(s)      \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) X(s)+\frac{1}{2}O_2(g)\rightarrow XO(s)    \Delta H_1=-668.5kJ

(2) XCO_3(s)\rightarrow XO(s)+CO_2     \Delta H_2=+384.3kJ

The expression for enthalpy of the reaction follows:

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

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-668.5))+(1\times (-384.3))=-1052.8kJ

Hence, the \Delta H^o_{rxn} for the reaction is -1052.8 kJ.

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