2 AlCl3 (aq) + 3 Na2S (aq) --> 2 Al(OH)3 (s) + 6 NaCl (aq)
72.1 mL of 0.0792 M Na2S (aq) is added to a concentrated AlCl3 (aq) solution. What is the mass of the solid precipitate that forms? Round your final answer to three significant figures
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2 AlCl3 (aq) + 3 Na2S (aq) --> 2 Al(OH)3 (s) + 6 NaCl (aq) 72.1 mL of 0.0792 M Na2S (aq) is added to a concentrated AlCl3 (aq) solution. What is the mass of the solid precipitate that forms? Round your final answer to three significant figures
given: 2 NiCl3(aq) + 3 Na2S(aq) --> Ni2S3(s) + 6 NaCl(aq) If necessary use Mm (g/mol) = 165.04 78.05 213.59 If 70.0 mL of a 0.400 M NiCl3(aq) is mixed 40.0 mL of a 0.800 M Na2S(aq), how many moles of the excess reactant are leftover?
In a precipitation reaction between Al(NO3)3(aq) and Na2S(aq), 12.7 mL of 0.150 M Al(NO3)3(aq) completly reacted with 10.5 mL of Na2S(aq). What was the molarity of Na2S(aq)? 2 Al(NO3)3(aq) + 3 Na2S(aq) → Al2S3(s) + 6 NaNO3(aq) Note: Insert only the numerical value of your answer with three decimal places (do not include the units or chemical in your answer).
Al(OH)3(s)+3HCl(aq)→AlCl3(aq)+3H2O(l) 1. Calculate the number of grams of HCl that can react with 0.440 g of Al(OH)3. 2. Calculate the number of grams of AlCl3 formed when 0.440 g of Al(OH)3 reacts. 3. Calculate the number of grams of H2O formed when 0.440 g of Al(OH)3 reacts.
100. mL of 0.200 M AgNO3(aq) is mixed with 61 mL of 0.100 M Na2S(aq). Calculate the mass (in g) of any precipitate that is formed. Enter your answer to 2 decimal places or O if no precipitate forms.
100. mL of 0.200 M AgNO3(aq) is mixed with 30 mL of 0.100 M Na2S(aq). Calculate the mass (in g) of any precipitate that is formed. Enter your answer to 2 decimal places or if no precipitate forms.
2NaOH (aq) + MgCl; (aq) + 2NaCl (aq) + Mg(OH)2 (s),- how much 0.1 M NaOH solution will completely precipitate the Mg²+ in 0.8 L of 0.10 M MgCl, solution? Round to 2 significant figures. Provide your answer below: LNaOH FEEDBAC
A solution of NaCl(aq) is added slowly to a solution of lead nitrate, Pb(NO3)2(aq) Pb ( NO 3 ) 2 ( aq ) , until no further precipitation occurs. The precipitate is collected by filtration, dried, and weighed. A total of 10.47 g PbCl2(s) 10.47 g PbCl 2 ( s ) is obtained from 200.0 mL 200.0 mL of the original solution. Calculate the molarity of the Pb(NO3)2(aq) Pb ( NO 3 ) 2 ( aq ) solution. concentration: M
Calculate the number of milliliters of 0.664 M Ba(OH)2 required to precipitate all of the Al3+ ions in 154 mL of 0.408 M AlCl3 solution as Al(OH)3. The equation for the reaction is: 2AlCl3(aq) + 3Ba(OH)2(aq) 2Al(OH)3(s) + 3BaCl2(aq) _____mL Ba(OH)2
3. Pb(OH)2 (8) is amphoteric and forms the complex ion Pb(OH)/(aq) with excess OH in solution. Write the balanced equilibrium equation of Pb(OH)2 (s) in the presence of excess hydroxide. 4. A solution contains Pb(NO3)2 and Zn(NO3)2. Consult the solubility table and answer the following questions: a. NaOH (aq) is added to the solution. What precipitate (or precipitates) forms? b. Can you separate the two metal ions using NaOH? c. Nat (aq) is added to the solution. What precipitate (or...