Given the reaction: 3 Ba(NO): + 2 AIC, 2 Al(NO), + 3 BaC12- Ifyou wish to make 1

Question

Given the reaction: 3 Ba(NO): + 2 AIC, 2 Al(NO), + 3 BaC12- Ifyou wish to make 1.45 kg of BaCl2, what volume of a 3.552 M solution of AlCl3 would be required? 5. 6. Lead (II) nitrate reacts with ammonium phosphate to produce a precipitate. If 455 mL of a 2.55 M lead (II) nitrate solution is reacted with excess ammonium phosphate, what mass of the precipitate will be formed? Consider the reaction of Mg with HCl. If a long strip of Mg with a volume of 2.85 cm3 were allowed to react with excess HC, 7. A) How many moles of MgClz would be produced? B) What would be the molarity of MgCl2 if the total volume of the solution were 250 mL? 8. A sample of sandstone contains silica (SiO2) and calcite (CaCO;). When the sandstone is heated the CaCO3 decomposes by the reaction: CaCO3(s) CaO(s) + CO2(g). What is the percentage of silica in the sandstone if heating 18.7 mg of the rock yields 3.95 mg CO2?

Explanation / Answer

5.

Given the reaction;

3 Ba(NO3)2+2AlCl3 = 2Al(NO3)3 + 3 BaCl2

First calculate the mole of BaCl2:

Number of moles = amount in g / molar mass

= 1.45 kg*1000 g/1.00 kg / 208.23 g/mol

= 6.96 mole BaCl2

Now calculartt the moles of AlCl3, which is required to prepare 6.96 mole BaCl2;

6.96 mole BaCl2 * 2 mole AlCl3 / 3 mole BaCl2

= 4.64 mole AlCl3

We know that molarity = number of moles / volume in L

So volume in L = number of mole s/ molarity

= 4.64 mole AlCl3 /3.552 mole/ L

= 1.3069 L

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