Question

If 9.00 grams of Cr(ClO4)3 are mixed with 4.00 grams of K2CO3, how much reactant is left over?

Answer 1

K2CO3 + Cr(ClO4)3

this is double replacement:

Cr(CLO4)3 + K2CO3 = KCLO4 + Cr2(CO3)3

balanced:

2 Cr(CLO4)3 + 3 K2CO3 = 6 KCLO4 + Cr2(CO3)3

mol of  Cr(ClO4)3 = mass/MW = 9 / 350.3479 = 0.025688

mol of K2CO3 = mass/MW = 4 /284.0189 = 0.0140835

then

ratio is

2:3

0.025688 need 3/2

1.5*0.025688 = 0.038532 mol of K2CO3 bu we only have 0.0140835

thjerefore

this ilimiting

0.0140835*2/3 = 0.009389mol of CrClo43 reacted

then

0.025688-0.009389 = 0.016299 mol of f Cr(ClO4)3 mol

mass = mol*MW = 350.3479*0.016299 = 5.71032 g