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Pavel [41]
3 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]3 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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Answer: The molar mass of the gas is 9.878 g/mol.

Explanation:

According to Graham's law, the rate of diffusion is inversely proportional to square root of molar mass of gas.

Rate = \frac{1}{\sqrt{M}}

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M = molar mass of gas

As given gas diffuses 1/7 times faster than hydrogen gas. So, its molar mass is calculated as follows.

\frac{R_{1}}{R_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}\\

where,

M_{1} = molar mass of hydrogen gas

M_{2} = molar mass of another given gas

R_{1} = rate of diffusion of hydrogen

R_{2} = rate of diffusion of another given gas = \frac{1}{7}R_{1}

Substitute the values into above formula as follows.

\frac{R_{1}}{R_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}\\\frac{R_{1}}{\frac{1}{7}R_{1}} =  \sqrt{\frac{M_{2}}{2}}\\7 \times 1.414 = M_{2}\\M_{2} = 9.878 g/mol

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