Chemistry Homework Help Please!!!?

Question

The Reaction is as follows:

CaI2(aq)+ K2C2O4(aq) --> CaC2O4(s) + 2KI(aq)

a.) If you had .05 moles of each reactant, how many moles of each product could you make?

b.) If you had 10.0 grams of calcium iodide, how many grams of calcium oxalate could you potentiallly make?

c.) How many grams of potassium iodide could you make with 10.0 grams of calcium iodide?

Answer 1

a. The ratio of substances in the balanced equation is 1:1:1:2. So, if you have 0.05 moles of each reactant, you could form 0.05 mol of CaC2O4(s) and 0.10 mol of KI.

b. To calculate this, use the molar mass of CaI2 to convert 10.0 grams of CaI2 into moles. Since the ration of CaI2 to CaC2O4 is 1:1, you could form the same number of moles of CaC2O4. THen use the molar mass of CaC2O4 to convert that number of moles to grams.

c. Use the same approach as in b), but recognize that you can form twice as many moles of KI as you have CaI2.

Answer 2

according to the balanced reaction equation, the mole ratios are as follows;

CaI2:CaC2O4 --- 1:1
CaI2:KI --- 1:2
K2C2O4: CaC2O4 --- 1:1
K2C2O4:KI --- 1:1


1) thus, if there was .05 moles of each reactant, there would be 0.05mole of CaC2O4 and 2 x n(reactant) = 2 x 0.05 = 0.1 moles of KI. (n = moles)

2) according to the balanced equation, the mole ratio of calcium iodide and calcium oxalate is 1:1

use the formula;

n = m / M
n = moles
m = mass (g)
M = molar mass (g mol^-1)

n(CaI2) = m(CaI2) / M(CaI2)
n(CaI2) = 10g / (40+ 127+127)gmol^-1
n(CaI2) = 0.048309179moles

thus due to the mole ratio of 1:1,
n(CaC2O4) = 0.048309179moles
thus,
m(CaC2O4) = n(CaC2O4) x M(CaC2O4)
m(CaC2O4) = 0.048309179moles x (40+12+12+(16x4))gmol^-1
m(CaC2O4) = 6.183574879 g

thus, 6.183574879 g ofcalcium oxalate could be potentiallly made.

c) the mole ratio of KI and CaI2 is 2:1
m(CaI2) = 10g
n(CaI2) = 10g / (40+ 127+127)gmol^-1
n(CaI2) = 0.048309179moles

thus n(KI) = 0.048309179moles
As a result, m(KI) = 0.048309179moles x (39+127)gmol^-1

m(KI) = 8.019323671g

thus, 8.019323671g of KI could be made with 10.0 grams of calcium iodide.

hope this helps:-)

Answer 3

i have no idea.
just thought id answer this to tell you that im havin soo much trouble w/my hw right now. so u dont feel alone .. lol.