I\'m stuck. I need step by step so I can understand where and how you came to th
ID: 1029149 • Letter: I
Question
I'm stuck. I need step by step so I can understand where and how you came to the answer. Thanks!
22) Consider the reaction if 33.2 g of K2C03 is produced from reacting 40.0 g K02 with 25.0L 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? 5lo &.moles oe Coa b. How much K2CO3 can be produced from the 33.2 g of KO2 that is available? 23) Calculate the hydronium ion concentration in an aqueous solution with a pH of 7.8 at 25°C. (2 points)Explanation / Answer
22)
a) According to avogadro's law at STP 1 mole of gas occupy volume = 22.414 L
then 25 L CO2 at STP = 25 / 22.414 = 1.115 mole
mole of CO2 present = 1.115 mole
b) molar mass of KO2 = 71.1 g/mol
no.of mole of compound = mass of compound in gm / molar mass of compound
no.of mole of KO2 = 33.2 / 71.1 = 0.46695 mole
According to balanced chemical reaction 4 mole of KO2 produce 2 mole of K2CO3 molar retio of KO2 to K2CO3 is 2:1 proportion therefore 0.46695 mole of KO2 produce K2CO3 = 0.46695 / 2 = 0.23347 mole
0.23347 mole of K2CO3 produced
molar mass of K2CO3 = 138.21 g/mol
gm of compound = no. of mole X molar mass
gm K2CO3 produced = 0.23347 X 138.21 = 32.27 gm
32.27 gm of K2CO3 produced
23 )
[H3O+] = 10-pH
= 10-7.8
= 1.585 X 10-8 M
Hydonium ion concentration of aquestion soltuion with pH 7.8 = 1.585 X 10-12 M