Answer:
The maximum mass of water that could be produced by the chemical reaction is 2.6 grams.
Explanation:
[tex]HBr(aq)+NaOH(s)\rightarrow NaBr(aq)+H_2O(l)[/tex]
Moles of HBr = [tex]\frac{20.2 g}{81 g/mol}=0.2494 mol[/tex]
Moles of sodium NaOH = [tex]\frac{5.7 g}{40 g/mol}=0.1425 mol[/tex]
According to reaction 1 mol of HBr is reacting with 1 mol of NaOH.
Then 0.2494 moles of HBr will react with :
[tex]\frac{1}{1}\times 0.2494 mol=0.2494 mol[/tex] of NaOH.
According to reaction 1 mol of NaOH is reacting with 1 mol of HBr.
Then 0.1425 moles of sodium hydroxide will react with :
[tex]\frac{1}{1}\times 0.1425 mol=0.1425 mol[/tex] of HBr.
As we can see that NaOH is in limiting amount of moles. So, the amount of water formed will depend upon the amount of NaOH.
According to reaction, 1 mole NaOH gives 1 mole of water.
Then 0.1425 moles of NaOH will give:
[tex]\frac{1}{1}\times 0.1425 mol=0.1425 mol[/tex] of water.
Mass of the 0.1425 mol of water:
0.1425 mol × 18 g/mol = 2.565 g ≈ 2.6 g.
The maximum mass of water that could be produced by the chemical reaction is 2.6 grams.
2.5 g of water is produced.
The equation of the reaction is;
NaOH(aq) + HBr(aq) -----> NaBr(aq) + H2O(l)
Number of moles of NaOH = 5.7 g/40g/mol = 0.14 moles
Number of moles of HBr= 20.2 g/81g/mol = 0.25 moles
Since the reaction is in a 1:1 ratio, NaOH is the limiting reactant.
0.14 moles of NaOH yields 0.14 moles water
Mass of water produced = mass × molar mass
Molar mass of water = 18 g/mol
Mass of water produced = 0.14 moles water × 18 g/mol
= 2.5 g of water.
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