1) Determine the grams of the cation present in each of the following compounds.

Question

1) Determine the grams of the cation present in each of the following compounds. Show all work and place your answers in the space provided. i)

(1) 3.45 grams of Ca3(PO4)2

(2) 28.2 grams of Sr(OH)2

(3) 142.1 grams of NH4NO3

(2) Calculate the density (in g/L) of NO2 (g) at -38.1°C and 719. torr. Assume that the gas is acting ideal and

report your answer in the space provided.

(3) If 2.34 grams of ethane (C2H6) is combusted with 85.8 grams of molecular oxygen (O2), how many

grams of water will be produced? Report your answer using the correct number of sig figs in the

space provided.

(4) How many grams of sodium hydroxide need to be added to 15.0 milliliters of a 1.24 molar sodium

hydroxide solution in order to produce a 15.0 milliliters of a 2.34 molar sodium hydroxide solution?

Assume that the volume of the added sodium hydroxide is negligible and report your answer with the

correct sig figs in the space provided.

(5) A sample of copper(II) sulfate pentahydrate (CuSO4 5H2O, MM = 249.71 g/mol) contains 3.59 * 1024

oxygen atoms. Determine how many grams of copper(II) sulfate pentahydrate are present in the

sample and report your answer in decimal notation using the correct number of sig figs in the space

provided.

(6) The combustion of butene (C4H8 , MM = 56.12 g/mol) in the presence of molecular oxygen (O2, MM =

32.00 g/mol) produces carbon dioxide (CO2 , MM = 44.01 g/mol) and water (H2O, MM 18.02 g/mol)

according to the following balanced equation:

C4H8(g) + 6 O2(g) 4 CO2(g) + 4 H2O(l)

If 38.5 grams of butene are burned with excess molecular oxygen, what is the percent yield of the

reaction if 113 grams of carbon dioxide are collected in lab? Write your answer using the correct

number of sig figs in the box provided.

Please do it all of them hear ((DO NOT WRITE IN A PAPER))

Explanation / Answer

Solution:- 1)

(1) we use stoichiometry for this kind of problems..

mass of compound -----> moles of compound ---------> moles of metal cattion --------> grams of metal cation

Ca3(PO4)2 ----------> 3Ca+2 + 2PO4-3

there is 1:3 mol ratio between compound and cation. Molar mass of compound, Ca3(PO4)2 is 310.18 g/mol and molar mass of Ca+2 is 40.078 g/mol.

3.45 g of Ca3(PO4)2 x (1mol of Ca3(PO4)2/310.18 g of Ca3(PO4)2) x (3 mol of Ca+2/ 1mol of Ca3(PO4)2) x (40.078 g of Ca+2/1mol of Ca+2) = 1.34 g of Ca+2

(2) Sr(OH)2 -------> Sr+2 + 2OH-

molar mass of Sr(OH)2 is 121.63 g/mol and that of Sr is 87.62 g/mol.

28.2 g of Sr(OH)2 x (1mol of Sr(OH)2/121.63 g of Sr(OH)2) x (1 mol of Sr+2/ 1 mol of Sr(OH)2) x (87.62 g of Sr+2/1 mol of Sr+2) = 20.3 g of Sr+2

(3) NH4NO3 ---------> NH4+ + NO3-

molar mass of NH4NO3 is 80.04 g/mol and that of NH4+ is 18.04 g/mol.

142.1 g of NH4NO3 x (1mol of NH4NO3/ 80.04 g of NH4NO3) x (1 mol of NH4+/1mol of NH4NO3) x (18.04 g of NH4+/ 1mol of NH4+) = 32.03 g of NH4+

2) If the gas behave ideally then,   d = PM/RT

where, d is density, P is pressure in atm, M is molar mass in g/mol. R is gas consstant with its value 0.0821 atm.L.mol-1.K-1 and T is kelvin temperature.

P is given = 719 torr x (1atm/760 torr) = 0.946 atm

T = -38.1 + 273 = 234.9 = 235 K

molar mass, M of NO2 = 46.0 g.mol-1

Let's plug in the values iin the formula...

d = 0.946 atm x 46.0 g.mol-1 /( 0.0821 atm.L.mol-1.K-1 x 235K)

d = 2.26 g/L

3) It is a stoichiometry problem so first of all we write the balanced equation..

2C2H6 + 7O2 --------> 4CO2 + 6H2O

we would calculate the grams of water for each of them and the correct theoretical yield woudl be the least one we get as that's the amount we get from limiting reactant.

2.34 g of C2H6 x (1mol of C2H6/30.07 g of C2H6) x ( 6 mol of H2O/ 2 mol of C2H6) x (18.02 g of H2O/1mol of H2O)    = 4.21 g of H2O

85.8 g of O2 x (1 mol of O2/32.0 g of O2) x ( 6 mol of H2O/ 7 mol of O2) x (18.02 g of H2O/1mol of H2O)    = 41.4 g of H2O.

From calculation it is clear that limiting reactant is C2H6 and it gives 4.21 g of H2O.

4) moles = molarity x volume of solution in liter

Initial moles of NaOH = 15.0 ml x (1L/1000ml) x (1.24 mol/L) = 0.0186 mol

final moles of NaOH = 15.0 ml x (1L/1000ml) x (2.34 mol/L) = 0.0351 mol

extra moles of NaOH added = 0.0351 - 0.0186 = 0.0165 mol

molar mass of NaOH is 40.0 g/mol

so, 0.0165 mol x (40.0 g/mol) = 0.660 g of NaOH

So, we need to add 0.660 g of NaOH.

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