Chemistry, 18.12.2019 20:31 sheldavis

Problem page aqueous hydrobromic acid will react with solid sodium hydroxide to produce aqueous sodium bromide and liquid water . suppose 80. g of hydrobromic acid is mixed with 55.0 g of sodium hydroxide. calculate the minimum mass of hydrobromic acid that could be left over by the chemical reaction. be sure your answer has the correct number of significant digits.

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To produce 0.9888 moles of NaBr and 0.9888 moles of H2O, there will remain 0.3862 moles of NaOH. All the HBr will be consumed, no HBr will be left over by the chemical reaction

Explanation:

Step 1: Data given

Mass of hydrobromic acid = 80.00 grams

Mass of sodium hydroxide = 55.00 grams

Molar mass HBr = 80.91 g/mol

Molar mass NaOH = 40 g/mol

Step 2: The balanced equation

HBr(aq) + NaOH(s) → NaBr(aq) + H₂O(l)

Step 3: Calculate moles of HBr

Moles HBr = Mass HBr / molar mass HBr

Moles HBr = 80.00 grams / 80.91 g/mol

Moles HBr = 0.9888 moles

Step 4: Calculate moles NaOH

Moles NaOH = 55.00 grams / 40.00 g/mol

Moles NaOH = 1.375 moles

Step 5: Calculate limiting reactant

For 1 mol of HBr we need 1 mol of NaOH to produce 1 mol of NaBr and 1 mol of H2O

The limiting reactant is HBr. It will completely be consumed. (0.9888 moles). NaOH is in excess. There will react 0.9888 moles. There will remain 1.375 - 0.9888 = 0.3862 moles

This means to produce 0.9888 moles of NaBr and 0.9888 moles of H2O, there will remain 0.3862 moles of NaOH. All the HBr will be consumed, no HBr will be left over by the chemical reaction

Answer from: Quest

може ли да зареждате повече информация, моля

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.

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Problem page aqueous hydrobromic acid will react with solid sodium hydroxide to produce aqueous sodi...

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