How many grams of PbBr 2 will precipitate when excess KBr solution is added to 4

Question

How many grams of PbBr2 will precipitate when excess KBr solution is added to 42.0 mL of 0.561MPb(NO3)2 solution?
Pb(NO3)2(aq) + 2KBr(aq) ---> PbBr2(s) + 2KNO3(aq)

???Grams

How many mL of 0.570 M HNO3 are needed to dissolve 5.90 g of MgCO3?
2HNO3(aq) + MgCO3(s) ----> Mg(NO3)2(aq) + H2O(l) + CO2(g)

??mL

Calculate the number of milliliters of 0.483 M KOH required to precipitate all of the Cr3+ ions in 143 mL of0.700 M CrI3 solution as Cr(OH)3. The equation for the reaction is:
CrI3(aq) + 3KOH(aq) ---> Cr(OH)3(s) + 3KI(aq)

???mL KOH

Explanation / Answer

Balanced equation:
Pb(NO3)2(aq) + 2 KBr(aq) ===> PbBr2(s) + 2 KNO3(aq)
Reaction type: double replacement

42.0 mL of 0.561MPb(NO3)2 solution = 42 x 0.561 /1000 = 0.023562 Moles

Moles of PbBr2 =  0.023562 Moles

Mass of PbBr2 =  0.023562 x 367.008 = 8.647 gm

Question 2

Balanced equation
2 HNO3(aq) + MgCO3(s) ===> Mg(NO3)2(aq) + H2O(l) + CO2(g)

5.90 g of MgCO3 = 5.9 / 84.31 = 0.0699 Moles

Moles of HNO3 needed =  0.13995 Moles (2 equivalent )

Volume of 0.57 M HNO3 = 0.13995 x 1000 / 0.57 = 245.52 ml

Question 3

Balanced equation:
CrI3(aq) + 3 KOH(aq) = Cr(OH)3(s) + 3 KI(aq)
Reaction type: double replacement
143 mL of0.700 M CrI3 solution = 143 x 0.7 / 1000 = 0.1001 Moles

Moles of KOH needed = 0.1001 x 3 = 0.3003 Moles

Volume of KOH needed = 0.3003 x 1000 / 0.483 = 621.739 ml

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