Molarity and Stoichiometry problems. Find the concentration of each ion present

Question

Molarity and Stoichiometry problems.

Find the concentration of each ion present in the solution formed by adding together the following: 45. 00 mL of 0. 1174 M K_2SO_4 and 35. 00 mL of 0. 2504 M HNO_3; 20. 00 mL of 0. 1025 M Cu(NO_3)_2 and 100. 00 mL of 0. 1055 M Fe(NH_4)_2(SO_4)_2. Assume all volumes are additive. What mass in grams of NaCl would be required to bring 4. 60 L of 0. 1180 M NaCl solution to a molarity of 0. 1225, without changing the volume of the solution? Find the following: How many grams of solute are present in 50. 0 mL of 0. 488 M K_2Cr_2O_7? If 4. 00 g of (NH_4)_2SO_4 are dissolved in enough water to make 400. 0 mL of solution, what is the molarity of that solution? How many mL of 0. 0250 M CuSO_4 contain 1. 75 g of solute? A 1. 000-g sample of limestone rock is crushed to powder and then treated with 50. 00 mL of 0. 1012 M HC1. The excess acid required 25. 26 mL of 0. 1050 M NaOH for neutralization. Calculate the percent by mass of the calcium carbonate in the rock. For the addition of 125. 00 mL of 0. 1352 M calcium bromide to 175. 00 mL of 0. 1015 M sodium oxalate, determine the following: Write the balanced molecular equation for the reaction. What is the limiting reagent? What is the molarity of all ions in the final solution? Assuming the reaction proceeds at 100 %, what volume of the limiting reagent is required to produce 45. 50 g of the precipitate if the concentrations remain the same? What molarity of the limiting reagent would be required if 100. 00 mL of that solution were used, and the desired amount of precipitate was 75. 00 g?

Explanation / Answer

Answer 2.

From definition of molarity we get,

0.118 M NaCl, 4.6L 58.44*0.118*4.6/1 = 31.720 g of NaCl.

(Molar mass of NaCl = 58.44 g/mol)

And 0.1225 M NaCl, 4.6L 58.44*0.1225*4.6/1 = 32.930 g of NaCl

So, additional amount of sodium chloride required = 1.210 g of NaCl

Answer 3(a)

            From definition of molarity we get,

0.488 M K2Cr2O7, 50 mL 294.185*0.488*50/1000 = 7.18 g (Molar mass of K2Cr2O7= 294.185 g/mol)

3(b) Molar mass of (NH4)2SO4 =132.14 g/mol

Number of moles = 4/132.14 = 0.030

400 ml contains = 0.030 mol

So, 1000 ml will contain = 0.030*1000/400 = 0.075 moles

Thus molarity = 0.075 M

3(c) Molar mass of CuSO4 = 159.609 g/mol

Molarity =0.025 M

Quantity = 1.75g

Volume = ?

By definition molarity, if the volume is V mL then,

V will contain 159.609*0.025*V/1000 = 1.75

Or V =1.75*1000/159.609/0.025 = 438 mL

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