[jbpoon]

A 10mL volume of 0.1mol/L of CuSO4 is reacted with 10mL of 0.15mol/L Na2CO3. What mass of CuCO3 precipiate is formed? answer below. Can you guys please check if it is correct. thanks.

CuSO4(aq)          +             Na2CO3                      CuCO3(s)          +          Na2SO4(aq)
V = 10mL                           V=10mL                         M=123.56g/mol
c = 0.1mol/L                   c = 0.15mol/L      

n = cV                          n = cV                                m = nM
V = 0.01L                      V = 0.01L                              m = (0.001mol)(123.56g/mol)
n = (0.01L)(0.1mol/L)            n = (0.01L)(0.15)                   m = 0.12356 g
n = 0.001 mol                n = 0.0015 mol

therefore the mass of CuCO3(s) is 0.12356g.

is this correct

[Borek]

See the image. Red means excess.

[jbpoon]

See the image. Red means excess.

ok... so based on that image my answer is correct. yes?

[Borek]

Looks like.

Just remember about significant digits.

[jbpoon]

Looks like.

Just remember about significant digits.

all right, thanks bud