To solve this problem, the dilution equation (M1 x V1 = M2 X V2) must be used. The given values in the problem are M1= 12.0 M, V1= 30 mL, and M2= 0.160 M. To solve for V2,
V2=M1xV1/M2
V2= (12x30)/0.16
V2= 2,250 mL.
The correct answer is 2.25 L.
Answer:
The correct answer is the first option
Explanation: hope this helped
The balanced reaction is 2KClO3 --> 2KCl + 3O2
We first divide the 400.0 g KClO3 by the molar mass of 122.55 g/mol to get 3.26 mol KClO3. Next, we use the coefficients: 3.26 mol KClO3 * (3 mol O2 / 2 mol KClO3) = 4.896 mol O2. Multiplying this by the molar mass of 32 g/mol gives 156.67 g O2.
Percent yield = 115.0 g / 156.67 g = 0.734 = 73.40%
First calculate the number of moles of each ion present
for Na2So4 is
0.52/142=0.0037moles
Na+ =0.0037 x2 =0.0074moles
SO4^-2 =0.0027moes
for Na3PO4= 1.186/164=0.0072 moles
Na+ =0.0072 x3=0.0216moles
po4 ions = 0.0072 moles
for LiSo4 =0.223/110=0.002moles
Li ions=0.002 x2 =0.004mole
SO4 ions =0.002moles
total moles for
Na ions =0.0216 +0.0074 =0.029moles
forSO4 ions =0.0037 +0.002=0.0057moles
molarity is therefore
Na ion=0.029/0.1=0.29M
SO4 ions=0.0057/0.1=0.057M
Li ions =0.004/0.1=0.04M
PO4 ions=0.0072/0.1=0.072M
