Question

What is the maximum amount of Li2CO3 that can form when 3.13 mol of CO2 reacts with excess LiOH?
CO2+2LiOH yields Li2CO3+H2O

A. 1.565 mol B. 3.13 mol
C. 6.26 mol D. 12.52 mol

Answer 1

B

Answer 2

how?

Answer 3

Please help me understand.

Answer 4

the soichiometry of the reaction is telling you that for each mole of CO2 you will produce one mole of Li2CO3, then if you have 3.13 moles of CO2 you will produce as a maximum of the same amount of moles 3.13 moles of Li2CO3. If the reaction were different and the coefficients of the two compounds were different you have to do the calculations according to the coefficients given in the reaction . If you have other problem let me know

Answer 5

thanks bro! So for each one produced on one side, then there is the same amount produced on the other side?

Answer 6

for that particular reaction. Other reaction can have other different relationships. You have to look at the coefficients. The coefficients are the numbers before the formula of the compound, if there is not number means that is 1.

Answer 7

If they were asking you about the LiOH the calculation is different. In this case you will need 2 moles of LiOH to produce one mole of Li2CO3

Answer 8

Oooo. Thank you.

Answer 9

Could you help me with a couple more? they're different.

Answer 10

Using the formula below. determine how many grams of oxygen are required to react with 10.0 g of magnesium. The molar mass of O2 is 32 g/mol, and the molar mass of Mg is 24.3 g/mol.
2Mg+O2 yields 2MgO

A. 2.2 g B. 3.3 g
C. 6.6 g D. 13.2 g

Answer 11

In this case you have to see that 2 moles of Mg will react with one moles of O2. If one mole of Mg is 24.3g, two mole of Mg will be 24.3 x 2 = 48.6g Mg
according to the reaction
48.6g Mg will react with 32 g O2, then 10gMg will react with 32/48.6g Mg =10x32/48.6 = 6.58gO2