Chemistry, 22.09.2019 21:30 laurenec49

If 22.99 g of sodium and 35.45 g of chlorine fully react, how much sodium chloride forms?

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Answer from: shamayajohnsonsh5

The amount of sodium chloride formed is 58.44 grams.

Given:

22.99 g of sodium completely reacts with 35.45 g of chlorine.

To find:

The mass of sodium chloride formed

Solution:

Mass of sodium = 22.99 g

Mass of chlorine = 35.45 g

The equation of chemical reaction between sodium and chlorine:

$2Na+Cl_2\rightarrow 2NaCl$

According to the question, 22.99 grams of sodium completely reacts with 35.45 grams of chlorine which means no reactants were left unreacted.

The mass of reactants = 22.99 g + 35.45 g = 58.44 g

The mass of product = Mass of NaCl formed =?

According to the law of conservation of mass:

Mass of reactants = Mass of products

Mass of reactants = 58.44 g

Mass of products = Mass of reactants = 58.44 g

Mass of sodium chloride formed = 58.44 g

The amount of sodium chloride formed is 58.44 grams.

Learn more about the law of conservation of the mass:

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Answer from: alexander1093

The answer to your question is: HCl = 58.44 g

Explanation:

To solve this problem, we must remember the law of conservation of mass that says that, matter cannot be created or destroyed, only changes its form.

That means that, the mass of the reactants must be equal to the mass of the products.

Process

Sodium (Na) + Chlorine (Cl) ⇒ Sodium chloride (HCl)

22.99 g 35.45 g x

22.99 + 35.45 = HCl

HCl = 58.44 g

Answer from: Quest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.

Answer from: Quest

The answer is c. h+ and oh− react to form water (h2o). i just took the test and i got it right. hope this !
Why is it impossible for a solution to become both more basic and more acidic at the same time?

Why is it impossible for a solution to become both more basic and more acidic at the same time?

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