Answer to Question #346226 in General Chemistry for Tebogo Ngobeni

Question

AH-453 kJ/mol-rxn P4010(S) + 6H₂O(1)-4H3PO4(aq) How many grams of P4010(S) would have to react to produce 167 kJ of energy?

Answer 1

M_r (P₄O₁₀) = 284 g/mol

P₄O₁₀ + 6 H₂O = 4 H₃PO₄ + 453 kJ

According to the above reaction, 453 kJ of energy is produced when 284 g of P₄O₁₀ reacts. We find how many grams of P₄O₁₀ need to react to produce 167 kJ of energy:

284 g P₄O₁₀ —> 453 kJ

X —> 167 kJ

X = 284 * 167 / 453 = 104,7 g

ANSWER: 104,7 g