In the reaction below, how many molecules of water can be produced by starting with 40.3 kilograms of sodium hydroxide and excess chlorine?
3Cl2 + 6NaOH -> 5NaCl + NaClO3+3H2O
a. 1.82 × 1027 molecules water
b. 1.01 × 1026 molecules water
c. 3.03 × 1026 molecules water
d. 6.07 × 1026 molecules water
e. None of these
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First, we need to determine the number of moles of sodium hydroxide (NaOH) present in 40.3 kilograms.
The molar mass of sodium hydroxide is 22.99 g/mol for sodium (Na), 16 g/mol for oxygen (O), and 1.01 g/mol for hydrogen (H).
Therefore, the molar mass of NaOH is 22.99 + 16 + 1.01 = 40 g/mol.
To convert from kilograms to grams, we multiply by 1000.
40.3 kilograms = 40,300 grams.
Now, we can determine the number of moles of NaOH:
Number of moles = mass / molar mass
Number of moles NaOH = 40,300 g / 40 g/mol = 1007.5 mol NaOH
According to the balanced chemical equation, the ratio of NaOH to H2O is 6:3, meaning for every 6 moles of NaOH, 3 moles of water are produced.
Therefore, we need to divide the number of moles of NaOH by 6 to find the number of moles of water:
Moles of water = 1007.5 mol NaOH / 6 = 167.9 mol H2O
Finally, we can convert the number of moles of water to molecules of water using Avogadro's number:
Number of water molecules = Number of moles of water × Avogadro's number
Number of water molecules = 167.9 mol H2O × 6.022 × 10^23 molecules/mol
Number of water molecules = 1.01 × 10^26 molecules.
The correct answer is b. 1.01 × 10^26 molecules of water.
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