In the reaction below, how many molecules of water can be produced from 40.7 g of sodium hydroxide and excess chlorine?
a. 1.23 × 1024 molecules water
b. 3.06 × 1023 molecules water
c. 1.53 × 1023 molecules water
d. 6.13 × 1023 molecules water
e. 8.17 × 1023 molecules water
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To determine the number of molecules of water produced, we first need to balance the equation for the reaction between sodium hydroxide and chlorine. The balanced equation is:
2 NaOH + Cl2 -> NaCl + H2O
From this equation, we can see that for every 2 moles (2 x 23g) of sodium hydroxide (NaOH), we will produce 1 mole (18g) of water (H2O).
First, we need to determine the number of moles of sodium hydroxide present in 40.7g. We can use the formula:
moles = mass / molar mass
The molar mass of sodium hydroxide (NaOH) is 23g + 16g + 1g = 40g/mol.
moles of NaOH = 40.7g / 40g/mol = 1.02 mol.
From the balanced equation, we can see that 2 moles of sodium hydroxide will produce 1 mole of water. Therefore, we can set up a proportion to determine the number of moles of water produced:
2 moles NaOH / 1 mole H2O = 1.02 moles NaOH / x
Cross-multiplying:
2 * x = 1.02
x = 1.02 / 2
x = 0.51 moles H2O
To convert moles of water to molecules, we can use Avogadro's number, which states that 1 mole of any substance contains 6.022 x 10^23 molecules.
Number of molecules H2O = 0.51 moles H2O * (6.022 x 10^23 molecules / 1 mole)
Number of molecules H2O = 3.06 x 10^23 molecules water
Therefore, the correct answer is b. 3.06 x 10^23 molecules water.
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