Answer to Question #74796 in Chemistry for johnny

Question

22) Consider the reaction if 33.2 g of K2CO3 is produced from reacting 40.0 g KO2 with 25.0 L of CO2 (at STP). The molar mass of KO2 = 71.10 g/mol and K2CO3 = 138.21 g/mol. (4 points)

4 KO2(s) + 2 CO2(g) → 2 K2CO3(s) + 3 O2(g)


a. How many moles of CO2 are available?






b. How much K2CO3 can be produced from the 33.2 g of KO2 that is available?

Answer 1

Answer:
a) n (CO2) = n (K2CO3) = 33.2/138.21 = 0.24 mol n = m/M
M (CO2) = 44 g/mol
m (CO2) = n·M = 0.24 · 44 = 10.56 g
b) n (K2CO3) = 1⁄2 n (K2O)
n (K2O) = 33.2/71.1 = 0.47 mol
n (K2CO3) = 0.47/2 = 0.235 mol
m (K2CO3) = 0.235 · 138.21 = 32.5 g