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Pavel [41]
2 years ago
9

Aqueous hydrobromic acid HBr will react with solid sodium hydroxide NaOH to produce aqueous sodium bromide NaBr and liquid water

H2O . Suppose 4.9 g of hydrobromic acid is mixed with 3.86 g of sodium hydroxide. Calculate the maximum mass of sodium bromide that could be produced by the chemical reaction. Round your answer to 2 significant digits.
Chemistry
1 answer:
Studentka2010 [4]2 years ago
5 0

Answer:

6.2g of NaBr are produced

Explanation:

The reaction of HBr with NaOH occurs as follows:

HBr + NaOH → NaBr + H2O

<em>Where 1 mole of each reactant produce 1 mole of NaBr</em>

To solve this question we need to find the moles of each reactant using their molar mass. With moles we can find limiting reactant and the moles (And mass) of NaBr produced, as follows:

<em>Moles HBr -Molar mass: 80.9119g/mol)-</em>

4.9g * (1mol/80.9119g) = 0.0606 moles HBr

<em>Moles NaOH -Molar mass: 40g/mol-</em>

3.86g * (1mol/40g) = 0.0965 moles NaOH

As the reaction is 1:1 and the moles of HBr < Moles NaOH, the limiting reactant is HBr and moles of NaBr produced are 0.0606 moles.

The mass of NaBr (Molar mass: 102.894g/mol) is:

0.0606 moles * (102.894g/mol) =

<h3>6.2g of NaBr are produced</h3>
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Acetaldehyde (CH3CHO) undergoes a Wolf-Kishner reaction, which is the addition of hydrazine (H2NNH2) with subsequent addition of
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Answer:

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Explanation:

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6 0
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Hydroxylamine is a weak molecular base with kb = 6.6 x 10-9. what is the ph of a 0.0500 m solution of hydroxylamine?
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Kb(NH₂OH) = 1,8·10⁻⁵<span>.
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c(NH</span>₂⁺) = c(OH⁻) = x.

c(NH₂OH<span>) = 0,05 mol/L - x.
Kb = c(NH</span>₂⁺) · c(OH⁻) / c(NH₂OH).

0,0000000066 = x² /  (0,05 mol/L - x). 

solve quadratic equation: x = c(OH⁻) = 0,000018 mol/L.<span>
pOH = -log(</span>0,000018 mol/L) = 4,74.<span>
pH = 14 - 4,74 = 9,23.</span>


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