Question

a) C₅H₁₂(l) + O₂(g) → CO₂(g) + H₂O(l)

If 28.6 grams of pentane (C₅H₁₂) are burned in excess oxygen, how many grams of H₂O will be produced?

____________g H₂O

b) If baking soda (sodium hydrogen carbonate) is heated strongly, the following reaction occurs:

2NaHCO₃(s) → Na₂CO₃(s) + H₂O(g) + CO₂(g)

Calculate the mass of sodium carbonate that will remain if a 1.28 g sample of sodium hydrogen carbonate is heated

___________  g Na₂CO₃

c) P₂O₅(s) + 3H₂O(l) → 2H₃PO₄(aq)

Given an initial mass of 10.35 g P₂O₅, an excess of H₂O (assuming that all of the reactant is converted to product(s), and none is lost), calculate the mass (g) of H₃PO₄ produced by the reaction.

__________ g

d)

"Smelling salts," which are used to revive someone who has fainted, typically contain ammonium carbonate, (NH₄)₂CO₃. Ammonium carbonate decomposes readily to form ammonia, carbon dioxide, and water. The strong odor of the ammonia usually restores consciousness in the person who has fainted. The unbalanced equation is

(NH₄)₂CO₃(s) → NH₃(g) + CO₂(g) + H₂O(g)

Calculate the mass of ammonia gas that is produced if 1.30 g of ammonium carbonate decomposes completely.

______ g NH₃

Answer 1

C₅H₁₂(l) +8 O₂(g) → 5CO₂(g) + 6H₂O(l)

1 mole of C5H12 combustion in presence of oxygen to gives 6 moles of H2O

72 gm of C5H12 combustion in presence of oxgen to gives 6*18gm of H2O

28.6 gm of C5H12 combustion in presence of oxygen to gives = 6*18*28.6/72 = 42.9 of H2O >>>> answer

b.

2NaHCO₃(s) → Na₂CO₃(s) + H₂O(g) + CO₂(g)

2 moles of NaHCO3 on heating to gives 1 mole of Na2CO3

2*84gm of NaHCO3 on heating to gives 106 gm of Na2CO3

1.28 gm of NaHCO3 on heating to gives = 106*1.28/2*84 = 0.8076gm of Na2CO3

c) P₂O₅(s) + 3H₂O(l) → 2H₃PO₄(aq)

    1 mole of P2O5 react with water to form 2 moles of H3PO4

   142 gm of P2O5 react with water to form 2*98gm of H3PO4

10.35 gm of P2O5 react with water to form = 2*98*10.35/142 = 14.28 gm of H3PO4 ...>>>> answer

d.

(NH₄)₂CO₃(s) → 2NH₃(g) + CO₂(g) + H₂O(g)

1 mole of (NH₄)₂CO₃ decoposes to gives 2 moles of NH3

96gm of (NH₄)₂CO₃ decomposes to gives 2*17gm of NH3

1.3gm of (NH₄)₂CO₃ decomposes to gives = 2*17*1.3/96   = 0.4604 gm of NH3