Question

Anhydrate = MgSO4(s)

Mass of container = 21.400g

Mass of container and content before heating = 26.65g

Mass of hydrate = 5.250

Mass of container and content after heating = 23.960g

Mass of anhydrate = 2.56g

Mass of water lost = 2.69g

Calculate moles of anhydrate in sample 2.56g anhydrate*1mole anhydrate/120.3626 molar mass= 0.0213 moles of anhydrate.

Calculate the ratio of moles of water for 1 mole of anhydrate: 0.149 calculated moles of H2O/0.0213 calculated moles of anyhydrate = 6.995305164/1mole anhydrate = 7 moles H2O/1 mole MgSO4

Use the whole number values calculated in above problem to write the formula for the hydrate =MgSO4(s0*7H2O (g)

Write and equation for the dehydration of the hydrate= MgSO4(s)*7H2O(g) MgSO4 (s) + 7H2O(g)

Write an equation for the dehydration of the following hydrates;

CaCl2*2H2O CaCl2*2H2O CaCl2 +2H2O

CoCl2*6H2O CoCl2*6H2O CoCl2 + 6H2O

Na2CO3*10H2O Na2CO3H2O Na2CO3H2O + 9H2O

Calculate the % water for the hydrate, Na2CO3*10H2O

10 *18/ 286 *100% =63% H2O in Na2CO3*10H2O

If you did not heat the hydrate sample sufficiently to drive off all of the water, would the % of water calculated be too high or too low? Explain.

Aluminum and oxygen gas react to produce aluminum oxide. What is the % of oxygen in the product aluminum oxide?

Answer = 89%

Write a balanced equation for the reaction = 4Al + 6O2 2Al2O3

If 75g of oxygen react, how many grams of Aluminum are required?

How many grams of aluminum oxide can be produced form 75g of oxygen?