Stoichiometry Find Molarity/moles of reactants or products?

[匿名]

2016-01-11 8:30 am

I just got back from winter break and I forgot a lot about Stiochiometry and I was asked these two questions that I have no idea how to do, could anyone help me?

1. If 21.5ml of a 1.63 M solution of MgCO3 reacts completely with 76 ml of a solution of Pb(NO3)2 what is the Molarity of the Pb(No3)2 solution

Rxn Equation: Pb(NO3)2 + MgCO3 > Mg(NO3)2 + PbCO3

And
2. How many grams of BaCrO4 will be precipitated when 24.2 ml of .65 M Ba(NO3)2 are treated with ab excess of Al2(CrO4)3

RXN equation: Al2(CrO4)3 + 3Ba(NO3)2 > 3BaCrO4 + 2Al)NO3)3

Any help will be appreciated!

回答 (1)

fooks

2016-01-11 9:45 am

1.
Molar mass of MgCO3 = 24.3 + 12 + 16×3 = 84.3 g/mol
Molar mass of Pb(NO3)2 = 207.2 + 14×2 + 16×6 = 331.2 g/mol

Pb(NO3)2 + MgCO3 → Mg(NO3)2 + PbCO3
1 mol of MgCO3 completely reacts with 1 mol of Pb(NO3)2.

No. of moles of MgCO3 = (1.63 mol/L) × (21.5/1000 L) = 0.03505 mol
No. of moles of Ph(NO3)2 = 0.03505 mol
Molarity of Pb(NO3)2 = (0.03505 mol) ÷ (76/1000) = 0.461 M

2.
Molar mass of BaCrO4 = 137.3 + 52 + 16×4 = 253.3 g/mol

Al2(CrO4)3 + 3Ba(NO3)2 > 3BaCrO4 + 2Al(NO3)3
1 mol of Ba(NO3)2 reacts to give 1 mol of BaCrO4.

No. of moles of Ba(NO3)2 reacted = (0.65 mol/L) × (24.2/1000 L) = 0.01573 mol
No. of moles of BaCrO4 formed = 0.01573 mol
Mass of BaCrO4 formed = (0.01573 mol) × (253.3 g/mol) = 3.98 g

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