Chapter 6: Quantities in Chemical Reactions page 244 question 6.21 SHOW WORK! ST

ID: 781987 • Letter: C

Question

Chapter 6: Quantities in Chemical Reactions

page 244 question 6.21

SHOW WORK! STEP-BY-STEP

When active metals such as sodium are exposed to air, they quickly form a coating of metal oxide.

The balanced equation for the reaction of sodium metal with oxygen gas is

4Na(s) + O2(g) ---------> 2Na2O(s)

Suppose a piece of sodium metal gains 2.05 g of mass after being exposed to air.

Assume that this gain can be attributed to its reaction with oxygen.

a) What mass of O2 reacted with the Na?

Answer: 2.05 g
b) What mass of Na reacted?

Answer: 5.89 g
c)What mass of Na2O formed?

Answer: 7.94 g

moles of Na = 2.05/23 = .089

so moles of O2 = .089/4 =.0222

wt = .0222*32=.713 gms

b) mass of Na = 2.05

c) mass of Na2O formed = .089*62/2 =2.759 gms

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