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When 50.0 mL of aqueous 0.306 M (NH4)2CO3 and 50.0 mL of aqueous 0.229 M AlCl3...

When 50.0 mL of aqueous 0.306 M (NH4)2CO3 and 50.0 mL of aqueous 0.229 M AlCl3 are mixed together, what is the mass in grams of solid formed? Show your work in the space below

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Expert Solution

3(NH4)2CO3 +2 AlCl3 ---------------- Al2(CO3)3(s) +6 NH4Cl

(NH4)2CO3 = 50.0ml of 0.306M

Number of moles of (NH4)2CO3 = 0.306M x 0.050L = 0.0153moles

AlCL3 = 50.0ml of 0.229 M

Number of moles of AlCl3 = 0.229Mx0.050L = 0.01145moles

According to equation , 2 moles of AlCL3 = 3 moles of (NH4)2CO3

                                       0.01145 moles of AlCL3 = ?

                                                                                = 3 x0.01145/2 = 0.017175 moles of (NH4)2CO3

we need 0.017175 moles of (NH4)2CO3. but we have 0.0153 moles of (NH4)2CO3. so it is used up first in the reaction. Hence it is limiting reagent.

A/C to equation , 3 moles of (NH4)2CO3 = 1 mole of Al2(CO3)3

                            0.0153 moles of (NH4)2CO3 = ?

                                                        = 1 x 0.0153/3 = 0.0051 moles of Al2(CO3)3

number of moles of Al2(CO3)3 formed = 0.0051 moles

molar massof Al2(CO3)3 =233.98gram/mole

mass of 0.0051 moles of Al2(CO3)3 = 233.98x0.0051 = 1.193grams.

mass of solid formed = 1.193 grams.

queen_honey_blossom answered 2 years ago
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