Answer:
A. 837 grams of [tex]H_3PO_4[/tex] are produced when 12.81 moles of water react with an excess of [tex]P_4O_{10}[/tex]
Explanation:
The balanced chemical equation is
[tex]P_4 O_10 + 6 H_2 O > 4 H_3 PO_4[/tex]
Mole ratio of [tex]H_2 O:H_3 PO_4[/tex] is 6 : 4 or 3 : 2
As per the Equation
6 moles of water produces 4 moles of Phosphoric acid
So let us convert
12.81 moles [tex]H_2 O[/tex] to moles [tex]H_3 PO_4[/tex] by using mole ratio and then to mass [tex]H_3 PO_4[/tex] by multiplying with molar mass [tex]H_3 PO_4[/tex]
[tex]12.82 mol H_2 O \times \frac {(4mol H_3 PO_4)}{(6molH_2 O)} \times \frac {(98gH_3 PO_4)}{(1mol H_3 PO_4 )}[/tex]
[tex]=837g H_3 PO_4[/tex] is produced
Please note:
Molar mass is the mass of 1 mole of the substance and its unit is g/mol
We find the molar mass by adding the atomic mass of the atoms present in it.
For example [tex]H_3 PO_4[/tex] contains 3 atoms of H, 1 atom of P and 4 atoms of O
So the molar mass
[tex]=(3\times1.008)+(31\times1)+(4\times16)=98 g/mol[/tex]
