30.0 g of P_4O_14 is mixed with 75.0 g of water to form phosphoric acid: P_4O_10

Question

30.0 g of P_4O_14 is mixed with 75.0 g of water to form phosphoric acid: P_4O_10 (s) + H_2 O (l) H_3 PO_4 (aq) a. Balance the equation. b. What is the limiting reactant? Show calculations. c. How many grams of phosphoric acid will form? Show calculations. Show work for full credit. Use Hess's Law to calculate the standard enthalpy for the reaction C_2 H_2 (g) + 2 H_2 (g) rightarrow C_2 H_6 (g) using the following combustion data: C_2 H_2 (g) + 5/2 O_2 (g) rightarrow 2CO_2 (g) + H_2 O (l) Delta H degree = -1300 kJ H_2 (g) + 1/2 O_2 (g) rightarrow H_2 O (I) Delta H degree = -286 kJ C_2 H_6 (g) + 7/2 O_2 (g) rightarrow 2CO_2 (g) + 3H_2 O (l) Delta H degree = -1560 kJ

Explanation / Answer

(a) Balanced equation is,

P4O10 + 6 H2O ---------> 4 H3PO4

(b)

Number of moles of P4O10 = mass / molar mass = 30.0 / 284. = 0.106 mol

Nmber of moles of water = 75.0 / 18 = 4.17 mol

From the balanced equation,

1 mol P4O10 needs 6 mol water

then, 0.106 mol P4O10 needs 6 * 0.106 = 0.636 mol of water ( < 4.17)

Hence P4O10 is limiting reagent.

(c)

From the balanced equation,

1 mol P4O10 forms 4 mol H3PO4

then, 0.106 mol P4O10 forms 4 * 0.106 = 0.424 mol H3PO4

Therefore,

Mass of Phosphoric acid formed = moles * molar mass = 0.424 * 98 = 41.6 g.

****

To get the final equation,

Do, (1) * 1 + (2) *2 - (3)*1

deltaH0rxn = - 1300 * 1 + (-286)*2 - ( - 1560 ) * 1 = - 312 kJ

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