14. The balanced equation below represents the reaction of KOH with H,PO4. коНаq) + Н,РО,(aq) a KH,PO,(aq) + H,O(1) When 20 grams of KOH (56 g/mol) reacts with 49 grams of H,PO, (98g/mol) How many grams of KH,PO, (136g/mol) are produced2 (Eirst Find L R)

Chemistry for Engineering Students
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Chapter4: Stoichiometry
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14. The balanced equation below represents the reaction of KOH with H3PO4. KOH(aq) + H3PO4(aq) à KH2PO4(aq) + H2O(l) When 20 grams of KOH (56 g/mol) reacts with 49 grams of H3PO4 (98g/mol) How many grams of KH2PO4 (136g/mol) are produced? (First Find LR)
14. The balanced equation below represents the reaction
of KOH with H,PO4.
КОН(aq) + H,РО,(аq) ӑ КН,РО,(аq) +
H,O(1)
When 20 grams of KOH (56 g/mol) reacts with 49
grams of H,PO4 (98g/mol)
How many grams of KH,PO, (136g/mol) are
produced? (First Find LR)
Transcribed Image Text:14. The balanced equation below represents the reaction of KOH with H,PO4. КОН(aq) + H,РО,(аq) ӑ КН,РО,(аq) + H,O(1) When 20 grams of KOH (56 g/mol) reacts with 49 grams of H,PO4 (98g/mol) How many grams of KH,PO, (136g/mol) are produced? (First Find LR)
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