Respuesta :
H3PO4 + 2 NH3 >> (NH4)2HPO4
moles NH3 = 10.00/ 17.3052 g/mol = 0.5872
moles fertilizer = 0.5872 / 2 = 0.2936
moles H3PO4 = 2800000 g / 98 g/mol = 28571
moles NH3 = 2 x 28571 =57142
mass NH3 = 57142 x 17.3052 = 988853 g => 989 Kg
moles NH3 = 10.00/ 17.3052 g/mol = 0.5872
moles fertilizer = 0.5872 / 2 = 0.2936
moles H3PO4 = 2800000 g / 98 g/mol = 28571
moles NH3 = 2 x 28571 =57142
mass NH3 = 57142 x 17.3052 = 988853 g => 989 Kg
Answer: 971.428 kg of ammonia will react with 2800 kg of phosphoric acid.
Explanation:
Mass of sodium hydrogen carbonate in 1 tablet =2800 kg = 2,800,000 g
[tex]H_3PO_4+ 2NH_3\rightarrow (NH_4)_2HPO_4[/tex]
Moles of [tex]H_3PO_4=\frac{\text{Mass of }H_3PO_4}{\text{Molar Mass of }H_3PO_4}=\frac{2,800,000 g}{98 g/mol}=28,571.42 moles[/tex]
According to reaction 1 mole of[tex]]H_3PO_4[/tex] reacts with 2 mole of [tex]NH_3[/tex] then 28,571.42 mole of[tex]H_3PO_4[/tex] will react with: [tex]\frac{2}{1}\times 28,571.42 [/tex] moles of [tex]NH_3[/tex] that is 57,142.84 moles.
Mass of [tex]NH_3=17 g/mol\times 57,142.84 moles.=971,428.28 g=971.428 kg[/tex]
971.428 kg of ammonia will react with 2800 kg of phosphoric acid.