Question

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1. What mass of Ca3(PO4)2 is produced if 750 mL of 6.00 M H3PO4 react according to the following reaction?
H3PO4 + Ca(OH)2 --> Ca3(PO4)2 + H2O




2. Using the reaction above, how many liters of 5.00 M H3PO4 are needed to produce 4390 g of Ca3(PO4)2?





3. If 75 mL of AgNO3 solution reacts with enough copper (II) to produce 0.25 g of Ag in a single displacement reaction, what is the molarity of the initial AgNO3 solution?





4. Suppose that 48 mL of 0.395 M Na2SO4 are added to Ca(NO3)2 to produce NaNO3 and CaSO4. Find the mass of CaSO4 produced.

Answer 1

Answer:

See Explanations

Explanation:

1. 2H₃PO₄(aq) + 3Ca(OH)₂(aq) => Ca₃(PO₄)₂(s) + 6H₂O(l)

100ml(6M H₃PO₄) + Excess Ca(OH)₂

=> 0.100(6)mole H₃PO₄ + Excess Ca(OH)₂

=> 0.600 mole H₃PO₄ + Excess Ca(OH)₂

from reaction stoichiometry, 2 moles H₃PO₄ in excess Ca(OH)₂ =>  1 mole  Ca₃(PO₄)₂

∴ 0.600 mole H₃PO₄ => 1/2(0.600 mole Ca₃(PO₄)₂ = 0.300 mole Ca₃(PO₄)₂

0.300 mole Ca₃(PO₄)₂ = 0.300 mole Ca₃(PO₄)₂ X 310.18 g/mole = 93.054 grams Ca₃(PO₄)₂(s)

2. 2H₃PO₄(aq) + 3Ca(OH)₂(aq) => Ca₃(PO₄)₂(s) + 6H₂O(l)

? Liters of 5M H₃PO₄ => 4390 grams Ca₃(PO₄)₂ = (4390g/310.18g/mol) 14.153 moles Ca₃(PO₄)₂

from equation stoichiometry, moles of H₃PO₄ used is 2 x moles Ca₃(PO₄)₂

Using equation =>  moles = Molarity X Volume

moles H₃PO₄ = (Molarity X Volume) H₃PO₂(aq) = 2 x moles Ca₃(PO₄)₂

∴ Volume H₃PO₂(in Liters) = 2 x 14.153 moles Ca₃(PO₄)₂ / 5.0M = 5.7 Liters of 5M H₃PO₄(aq) solution.

3. Cu°(s) + 2AgNO₃ => Cu(NO₃)₂ + 2Ag°

75ml of X-molar AgNO₃ => 0.25 gram Ag° = (0.25/108) mole Ag° = 0.0023 mole Ag°

(X-molar AgNO₃)(0.075 L) = 0.0023 mole Ag° => X-molar AgNO₃ = (0.0023 mole/0.075 Liter) = 0.031M AgNO₃(aq)

4.  48ml of 0.395M Na₂SO₄(aq) + Ca(NO₃)₂(aq) => 2NaNO₃(aq) + CaSO₄(s)

=> (0.048L)(03.395M)Na₂SO₄ = 0.019 mole Na₂SO₄ used = 0.019 mole CaSO₄(s) produced = (0.019 mole CaSO₄)(136.14 g/mole) = 2.58 grams CaSO₄(s) produced.