Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Question 6 Which of the following contain the most grams oflithium? o 0.1 oles L

ID: 1060815 • Letter: Q

Question

Question 6 Which of the following contain the most grams oflithium? o 0.1 oles Li2CO3 O 0.4 moles LiB O p moles LNO3 O 0.2 moles L13PO4 0.25 moles Li2so4 Question 8 e balanced equation for the combusion of ethane is 2 C2H6 7 02 4 CO2 6 H20 f 0.84 moles of water are in the reaction, how many moles of C2H6 were combuste (Remember to put a zero in front of the decimal point, like 0.55, for numbers less than one. so, remember to use the correct number of significant figures.) How many moles of oxygen are required for the production of 84 moles of water? If 0.84 moles of water are formed, how many moles of carbon dioxide are also formed? Question 9 Consider the reaction: 2 H2S 3 O2 2 SO2 2 H20

Explanation / Answer

Ans:

6.ans: Emperical formula of a compound with elemental ratio 1.00 mole of C to 2.67 mole of H.

To make it a whole number ratio, multiply by the simplest whole number.

C - 1.00 mole x 3 = 3.00 moles

H - 2.67 mole x 3 = 8.01 moles

Emperical formula will be C3H8

7.ans:

Option 1 is correct :

0.1 moles Li2CO3 contains the most grams of Li

Molecule(g) Li(g) mass %

option 1 - 7.381 1.3882 0.188077

option 2 - 34.78 5.5528 0.15965;

option 3 - 20.6838 2.0823 0.10067;

option 4 - 23.158 4.164 0.1798;

option 5 - 27.485 3.4705 0.1262

8.ans:

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

a.If 0.84 moles of H2O were formed in the reaction how many moles of C2H6 were combusted ?

moles of C2H6 combusted = 0.84 moles of H2O x 2 moles of C2H6

6 moles of H2O

= 0.28 moles C2H6

b. moles of O2 required = 0.84 moles of H2O x 7 moles of O2

6 moles of H2O

=   0.98 moles O2

c. moles of CO2 formed = 0.84 moles of H2O x 4 moles of CO2

6 moles of H2O

   = 0.56 moles of CO2

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote