Respuesta :
2.0 g NaSO4(1 mol NaSO4/142.04 g)(1 mol H2O/1 mol NaSO4)(18.02 g/1 mol H2O)(6.02x10^23 g/ 1 molecule H2O)= 1.5275 molecules of water
In the given reaction, the molecules of water produced are [tex]\bold{ 1.6\times10^2^2\; molecules }[/tex].
What are dehydration reactions?
The reactions in which water molecules are released are called dehydration reactions.
Step 1: first we will balance the equation
[tex]\bold{H_2SO_4 + 2NaOH = 2H_2O + Na_2SO_4 }[/tex]
The reaction tells us we produce 2 moles of [tex]\bold{H_2O}[/tex] for every 1 mole of [tex]\bold{Na_2SO_4}[/tex] produced.
Step 2: Now, if 2.0 moles of sodium sulfate is produced. That would mean we also had 4 moles of water, since 2 (moles H2O/molesNa2SO4).
[tex]\bold{n = \dfrac{n}{MM} = \dfrac{(2 g\; Na_2SO_4)}{(142)} = 0.014 \;mol\; Na_2SO_4}[/tex]
[tex]\bold{0.014\ mol\; NaSO_4 \times\dfrac{2H_2O}{1 Na S0_4} = 0.028\; mol\; H_2O}[/tex]
[tex]\bold{a = (n)(A) = (0.028 \;mol\; H_20)\times(6.02 \times 1023)= 1.6\times10^2^2\; molecules }[/tex]
Thus, the number of molecules of water produced in 2.0g of sodium sulfate = [tex]\bold{ 1.6\times10^2^2\; molecules }[/tex]
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