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KiRa [710]
3 years ago
5

How many milliliters of sodium metal, with a density of 0.97 g/mL, would be needed to produce 43.6 grams of sodium hydroxide in

the single replacement reaction below? Be sure to show the work that you did to solve this problem.
Chemistry
1 answer:
GaryK [48]3 years ago
7 0

Answer is: 25.84  milliliters of sodium metal.

Balanced chemical reaction: 2Na + 2H₂O → 2NaOH + H₂.

d(Na) = 0.97 g/mL; density of sodim.

m(NaOH) = 43.6 g; mass of sodium hydroxide.

n(NaOH) = m(NaOH) ÷ M(NaOH).

n(NaOH) = 43.6 g ÷ 40 g/mol.

n(NaOH) =1.09 mol; amount of sodium hydroxide.

From chemical reaction: n(NaOH) : n(Na) = 2 : 2 (1: 1).

n(Na) = 1.09 mol.

m(Na) = 1.09 mol · 23 g/mol.

m(Na) = 25.07 g; mass of sodium.

V(Na) = m(Na) ÷ d(Na).

V(Na) = 25.07 g ÷ 0.97 g/mL.

V(Na) = 25.84 mL.

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qwelly [4]

<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}=?

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

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