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
Explanation / Answer
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