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g Aqueous hydrobromic acid will react with solid sodium hydroxide to produce aqueous sodium bromide and liquid water . Suppose 57.4 g of hydrobromic acid is mixed with 41. 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.

Sagot :

Answer:

Zero

Explanation:

The complete reaction for this is as follows -

HBr + NaOH = NaBr + H2O

1 mole of NaOH reacts with 1 mole of HBr

m(HBr) = 57.4g and M(HBr) = 80.9g/mol

m(NaOH) = 16g and M(NaOH) = 40g/mol

Number of moles = m/M

Substituting the given values we get -

n(HBr) = 57.4 g/80.9 g/mol = 0.709

n(NaOH) = 16 g/40 g/mol = 0.4

n(H2O) = n(NaOH) = 0.40 mol

According to balanced equation

40 gram of NaOH reacts with 81 grams of HBr

41 gram of NaOH will react with 81/40 * 41 = 83.025 gram of HBr

We have only 57.4 g of HBr only hence, 0 HBr will be left.

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