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will then be the least quantity of calculated produc must correspond to the reactant that ran out first the limiting reactant. Example t as that -14 g of N2"were reacted with a g of Hn, what mass of NH 2 NH could form? of NH based on both the N and H. The right answer w the smaller calculated mas as thát must correspond to the reactant (be it N2 or H2) that ran out first (i.e te To solve this problem we wilí calculate the mass will be reactant). Thus the limiting Thusthe answer is 17g 2 How many moles of c coula form when 3.5 moles of A are reacted Additional stoichiometric examples: H2SO4 (aq) + 2 NaOH (aq).. 2 H2O + Na2 what volumé of 0.2 M H2S04 solution is neéded to"neütralize When volumes and molarities are involved the stoichiometric approach remains the same: get the moles of starting (now by M x V), use the mole ratio from the equation, and do inal conversion (often using molarity to convert back to : Note that molarity can be used as mmole/mL, as well as mole/L. Thus Co. 3 mmo 1 Na0H/m) x( 40mL) x (1 mmo 1r2 504/2mmo1 NaoH)xtimL/0.2mmo1r 2504) A1 (OH )2(S) 3 KNO (aq) + 3 KOH ( aq ) - + Al (NO3) 3 (aq) What mass of luminum hydroxide could form by the reactio of mI of 0.1 M Al(NO3 3 (aq) with an excess of 0.2 M KOH? y: 40 Ca( OH ) 2 ( ag) Ca3(PO 4 ) 2 (S),+ 6 H2O 3 c/10-S: H3 PO4 (aq) 2 + What mass of clcium phosphate could form by thereaction of 50 mL of 0.2 M H2PO, with 80 mL of 0.1 M Ca(OH)2 Note that this iš alimiting reactant problem so calculate the mass of Ca (PO4 based separately on the H3PO4 and on the Ca (OH)2, cñoosing the lesser calculated valde as the right answer..

Explanation / Answer

10.3

2A + 3B--------------- C + 4D

number of moles of A= 3.5 moles

number of moles of B= 4.8 moles

according to equation

2 moles of A = 3 moles of B

3,5 moles of A= ?

                        = 3x3.5/2 = 5.25 moles of B

we need 5,25 moles of B. but we have 4.8 moles of B.so B is completed first in the reaction.

Hence B is limiting reagent.

Accroding to equation

3 moles of B = 1 mole of C

4,8 moles of B = ?

                    = 4,8 x1/3 = 1.6 moles of C

number of moles of C formed = 1.6 moles.

10.4

Al(NO3)3    + 3 KOH ---------- Al(OH)3 (s)   +3 KNO3(aq)

Al(NO3)3 = 0.1M of 40ml

number of moles of Al(NO3)3 =0.1M x0.04L =0.004 moles

According to equation

1 mole of Al(NO3)3 = 1 mole of Al(OH)3

0.004 moles of Al(NO3)= ?

                              = 0.004 x 1/1 = 0.004 moles

Number of moles of Al(OH)3 = 0.004 moles

molar mass of Al(OH)3 = 78 gram/mole

mass of 0.004 moelsof Al(OH)3 = 0.004 x 78 = 0.312 grams

mass of Ai(OH)3 formed = 0.312 grams

10.5

2 H3PO4   + 3 Ca(OH)2 ------------   Ca3(PO4)2 + 6 H2O

H3PO4 = 0.2M of 50 ml

number of moles of H3PO4 = 0.2M x 0.05L= 0.01 moles

Ca(OH)2 = 0.1M of 80 ml

number of molesof ca(OH)2 = 0.1M x 0.08L = 0.008 moles

Accordign to equation

2 moles of H3PO4 = 3 moles of Ca(OH)2

0.01 moles of H3PO4 = ?

                                 = 0.01 x 3/2 = 0.015 moles of Ca(OH)2

we need 0.015 moles of Ca(OH)2. but we have 0.008 moles of Ca(OH)2. so Ca(OH)2 is completed in the reaction.

Hence Ca(OH)2 is limiting reagent

According to equation

2 moles of Ca(OH)2 = 1 mole of Ca3(PO4)2

0.008 moles of Ca(OH)2 = ?

                                  = 0.008 x 1/2 =0.004 moles of Ca3(PO4)2

number of molesof Ca3(PO4)2 = 0.004 moles

molar mass of Ca3(PO4)2 = 310.18 gram/mole

mass of 0.004 moles of Ca3(PO4)2 = 0.004 x 310.18 = 1.24072 grams

mass of Ca3(PO4)2 = 1.241 grams.

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