Answer to Question #181302 in General Chemistry for Joshua

Question

Solve the following word problem using the given handouts.

1) How many grams of calcium phosphate can be produced from the reaction of 2.50 L of 0.250 M Calcium chloride with and excess of phosphoric acid?  

 Calcium chloride + phosphoric acid --> calcium phosphate + hydrochloric acid 

 3CaCI2 + 2H3PO4 --> Ca3(PO4)2 + 6HCI

2) How many milliliters of 1.50 M Nitric acid is required to react with 100.0 g of cuprous oxide?

  14 HNO3 + 3Cu20 --> 6 Cu (NO3) 2 + NO + 7H2O

3) 60.5 mL of HNO3 are required to react with 25.0 mL of a 1.00 M Barium hydroxide solution:

   2 HNO3(aq) + Ba(OH)2 (aq) --> 2 H2O (s) + Ba(NO3)2 (aq) (BALANCED)

4) Calculate the grams of silver chloride produced from 10.00 ml of 10.0 M magnesium chloride with 100.0 ml of 2.20 M silver nitrate

   2AgNO3 + MgCI2 --> Mg ( NO3)2(s) + 2AgCI(aq)

Answer 1

1. Mole of CaCl2= 2.50 x 0.250= 0.625mol

From the balanced equation

3mol of CaCl2 yields 1mol of Ca3(PO4)2

0.625mol yields 1/3 x 0.625 = 0.208mol

Mass of Ca3(PO4)2 = 0.208x215= 44.72g

2. Molar mass of Cu2O= 143.08g/mol

Mole= 100.0/143.08= 0.7mol

From the balanced equation

3molof Cu2O reacts with 4mol of HNO3

0.7mol of Cu2O will react with 4/3 x 0.7= 0.93mol

Mole= cxV

V= mole/c = 0.93/0.250= 3.72L= 3720ml

3. Mole of Ba(OH)2 =1.00x25/1000 = 0.025mol

From the balanced equation

1mol of Ba(OH)2 reacts with 2mol HNO3

0.025mol of Ba(OH)2 will react with 0.050mol of HNO3

Mole= cxV

C= mole/v = 0.050/60.5 x 10^-3= 0.83M

4. Mole of MgCl2= 10.0 x 10.0/1000 = 0.10mol

Mole of AgNO3= 2.20 x 100.0/1000 = 0.22mol

Let's find the limiting reagent from the balanced equation

2mol of AgNO3 reacts with 1mol of MgCl2

0.22mol of AgNO3 should react with 0.11mol of MgCl2

But only 0.10mol of MgCl2 is available, hence it is the limiting reagent

Now,

1mol of MgCl2 gives 2mol of AgCl

0.10mol of MgCl2 will give 0.20mol of AgCl

Molar mass of AgCl= 108+35.5= 143.5g/mol

Mass of AgCl= 0.20x143.5= 28.7g