Chemistry with Lab, Lesson 105 answers

 

1) How many moles of sodium atoms correspond to 1.56×10^21 atoms of sodium? 
The number of moles is equal to the number of particles of a substance in a given sample divided by the Avogadro’s number.
Moles = Number of sodium atoms/ Avogadro’s number
Sodium atoms = 1.56 x 10^21
Avogadro number = 6.022 x 10^23
Moles = 1.56 x 10^21 / 6.022 x 10^23
Moles = 2.6 x 10^-3
So, there are 2.6 x 10^-3 moles of sodium in 1.56 x 10^23 atom.
2) How many moles of Al atoms are needed to combine with 1.58 mol of O atoms to make aluminum oxide, Al2O3?
4 Al + 3 O2 = 2 Al2O3
1.58 mol O2 x (4 mol Al/3 mol O2) = 2.11 mol Al

3) How many moles of Al are in 2.16 mol of Al2O3?

There are 2 Al atoms for each Al2O3 formula unit
2.16 mol Al2O3 x 2 = 4.32 mol Al

4) How many moles of H2 and N2 can be formed by the decomposition of 0.145 mol of ammonia, NH3?
2NH3 = 3H2 + N2
0.145 mol NH3 x (3 mol H2/2 mol NH3) = 0.218 mol H2
0.145 mol NH3 x (1 mol H2/2 mol NH3) = 0.0725 mol N2

5) What is the total number of atoms in 0.260 mol of glucose, C6H12O6?
There are 24 atoms in 1 mole of glucose. (6 carbon + 12 hydrogen + 6 oxygen = 24 total atoms)
0.260 mol C6H12O6 x (6.02 x 1023 molecules C6H12O6/1 mol C6H12O6) x (24 atoms/1 molecule C6H12O6) = 3.76 x 1024 atoms

6) Determine the mass in grams of each of the following:
a. 1.35 mol Fe 1.35 mol Fe x (55.845 g Fe/1 mol Fe) = 75.4 g Fe
b. 24.5 mol O 24.5 mols O x (15.999 g O/1 mol O) = 392 g O
c. 1.25 mol Ca3(PO4)2 1.25 mols x (310.174 g/1 mol) = 388 g Ca3(PO4)2
d. 0.625 mol Fe(NO3)3 0.625 mols x (241.857 g/1 mol) = 151 g Fe(NO3)3
e. 0.600 mol C4H10 0.600 mols C4H10 x (58.124 g C4H10/1 mol C4H10) = 34.9 g C4H10

7) Calculate the number of moles of each compound:
a. 21.5 g CaCO3 21.5g CaCO3 x (1 mol CaCO3/100.086 g CaCO3) = 0.215 mol CaCO3
b. 1.56 g NH3 1.56 g NH3 x (1 mol NH3/17.031 g NH3) = 0.0916 mol NH3
c. 16.8 g Sr(NO3)2 16.8 g x (1 mol/211.658 g) = 0.0794 mol Sr(NO3)2
d. 6.98 µg Na2CrO4d 6.98*10^-6 g x (1 mol/161.976 g) = 4.31×10^-8 moles
6.98 µg (micrograms) = 6.98×10^-6 grams

8) How many grams of O are combined with 7.14×1021 atoms of N in the compound N2O5?

There are 16g of Oxygen in one mole.
How many moles of Oxygen are in N2O5 if there are 7.14*10^21 atoms of Nitrogen? 7.14*10^21/6.022*10^23 = 1.19*10^-2 moles of Nitrogen
It’s a 2:5 ratio of Nitrogen to Oxygen, so (5/2)(1.19*10^-2) = 2.98*10-2 moles of Oxygen
16 grams/mole of Oxygen x 2.98*10^-2 moles = 0.4768 grams of Oxygen
(The answer says 0474g. That’s an acceptable difference due to rounding significant figures during the various calculations.)

9. When 0.684 g of an organic compound containing only C, H, and O was burned in oxygen 1.312g CO2 and 0.805g H2O were obtained. What is the empirical formula of the compound?
CO2 = 44 g/mol, H2O = 18 g/mol
Calculate grams of Carbon: 1.312g/(44 g/mol) = 0.029 mol C, 0.029 mol * 12 g/mol = 0.35g C
Calculate grams of Hydrogen: 0.805g/(18 g/mol) = 0.0447 * 2 = 0.089 mol H, 0.089 mol * 1 g/mol = 0.089g H
Calculate grams of Oxygen: 0.684 CxHxOx – (0.35g C + 0.089g H) = 0.245g O
Calculate moles of Oxygen: 0.245g/(16 g/mol) = 0.015 mol O
Figure out the ratio
C : H : O
0.029 : 0.089 : 0.015
2 : 6 : 1
C2H6O

10) Chlorine is used by textile manufacturers to bleach cloth. Excess chlorine is destroyed by its reaction with sodium thiosulfate.
a. How many moles of Na2S2O3 are needed to react with 0.12mol of Cl2?
0.12 mol Cl2 x (1 mol Na2S2O3/4 mol Cl2) = 0.030 mol Na2S2O3
b. How many moles of HCl can form from 0.12mol of Cl2?
0.12 mol Cl2 x (8 mol HCl/4 mol Cl2) = 0.24 mol HCl
c. How many moles of H2O are required for the reaction of 0.12mol of Cl2?
0.12 mol Cl2 x (5 mol H2O/4 mol Cl2) = 0.15 mol H2O
d. How many moles of H2O react if 0.24mol HCl is formed?
0.24 mol HCl x (5 mol H2O/8 mol HCl) = 0.15 mol H2O

11) The incandescent white of a fireworks display is caused by the reaction of phosphorous with O2 to give P4O10.
a. Write the balanced chemical equation for the reaction. P4 + 5O2 = P4O10
b. How many grams of O2 are needed to combine with 6.85g of P?
6.85 g P4 x (1 mol P4/123.896g P4) x (5 mol O2/1 mol P4) x (31.998gO2/1 mol O2) = 8.85 g O2
c. How many grams of P4O10 can be made from 8.00g of O2?
8.00 g O2 x (1 mol O2/31.998 g O2) x (1 mol P4O10/5 mol O2) x (283.886g P4O10/1 mol P4O10) = 14.2 g P4O10
d. How many grams of P are needed to make 7.46g P4O10?
7.46 g P4O10 x (1 mol P4O10/283.886 gP4O10) x (1 mol P4/1 mol P4O10) x (123.896 g P4/1 mol P4) = 3.26 g P4

12) In dilute nitric acid, HNO3, copper metal dissolves according to the [equation on your worksheet]. How many grams of HNO3 are needed to dissolve 11.45g of Cu? 
11.45 g Cu x (1 mol Cu/64.546 g Cu) x (8 mol HNO3/3 mol Cu) x (63.012 g HNO3/1 mol HNO3) = 30.28 g HNO3

13) The reaction of powdered aluminum and iron(II)oxide, 2Al(s) + Fe2O3(s) = Al2O3(s) + 2Fe(l) produces so much heat the iron that forms is molten. Because of this, railroads use the reaction to provide molten steel to weld steel rails together when laying track. Suppose that in one batch of reactants 4.20mol Al was mixed with 1.75mol Fe2O3.
a. Which reactant, if either, was the limiting reactant?
4.20 mol Al /2 = 2.10
1.75 mol Fe2O3 /1 = 1.75 mol = smaller number, LR
b. Calculate the mass of iron (in grams) that can be formed from this mixture of reactants.
1.75 mol Fe2O3 x (2 mol Fe/1 mol Fe2O3) x (55.845 g Fe/ 1 mol Fe) = 195 g Fe