25 mL of a 0.8 M solution of [tex]H_3PO_4[/tex] is reacted with [tex]NaOH[/tex] to produce [tex]H_2O[/tex] and [tex]Na_3PO_4[/tex]. The density of the solution is 1.55 g/mL. If all the phosphoric acid is allowed to react, what mass of sodium phosphate will be produced?

From the equation, we can see that 1 mole of [tex]H3PO4[/tex] reacts with 3 moles of NaOH to produce 1 mole of [tex]Na_3PO_4[/tex]. We can use this stoichiometry to calculate the amount of Na3PO4 produced from the given volume and concentration of [tex]H3PO4[/tex].

First, we need to calculate the number of moles of [tex]H_3PO_4[/tex] present in the solution:

moles [tex]H_3PO_4[/tex] = concentration x volume = 0.8 mol/L x 0.025 L = 0.02 moles

Next, we can use the stoichiometry of the reaction to determine the number of moles of [tex]Na_3PO_4[/tex] produced:

moles

[tex]Na_3PO_4[/tex] = 1/3 x moles [tex]H_3PO_4[/tex] = 1/3 x 0.02 moles = 0.0067 moles

Finally, we can calculate the mass of [tex]Na_3PO_4[/tex] produced using its molar mass and the number of moles calculated above:

[tex]Na_3PO_4[/tex] mass = moles [tex]Na_3PO_4[/tex] x molar mass = 0.0067 moles x 163.94 g/mol = 1.10 g

The mass of [tex]Na_3PO_4[/tex] produced is 1.10 g. We can also calculate the mass of the H3PO4 used in the reaction using its density:

mass H3PO4 = volume x density = 0.025 L x 1.55 g/mL = 0.0388 g

This confirms that all of the [tex]H_3PO_4[/tex] has reacted to form [tex]Na_3PO_4[/tex].

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