Hey there ! :
<span>2 Na + 2 H2O = 2 NaOH + 1 H<span>2
</span></span>Sodium<span> + </span>Dihydrogen Monoxide<span> = </span><span>Natriumhydroxid</span><span> + </span>Hydrogen
Coefficients:
Reagents : Na = 2
H2O = 2
Products : NaOH = 2
H2 = 1
POH will be -log[conc of OH]
-log (3.9E-08) = 7.409
pH = 14- pOH
pH = 14 - 7.409
pH = 6.59
1.27moles of K3PO4 can be formed.
<u>Answer:</u> The moles of water produced are 1.54 moles.
<u>Explanation:</u>
To calculate the number of moles, we use the equation:

Given mass of ethane = 15.42 g
Molar mass of ethane = 30.07 g/mol
Putting values in above equation, we get:

The chemical equation for the combustion of ethane follows:

By Stoichiometry of the reaction:
2 moles of ethane produces 6 moles of water
So, 0.513 moles of ethane will produce =
of water
Hence, the moles of water produced are 1.54 moles.