Given the unbalanced equation, Cr(Nobaa) + LioH(aq) ? many grams of Cr(OHla are

Question

Given the unbalanced equation, Cr(Nobaa) + LioH(aq) ? many grams of Cr(OHla are produced from 1.31 g of LiOH? uN01aq) + CriOH)|s), how Your Answer Answer Save Question 7 (1 point) Given the balanced equation, Na SOlag)+BaCag) 2 NaCl(ag) moles of Na:sO4 are required to produce 0.89 g of Baso, Your Answer BaSOls), how many Answer Save Question 8 (1 point) A chemical reaction is predicted to produce 105.7 grams of PbCro If 0.20 moles of PbCrO are experimentally obtained, what is the percent yield? Your Answer 9

Explanation / Answer

6)

Molar mass of LiOH,

MM = 1*MM(Li) + 1*MM(O) + 1*MM(H)

= 1*6.968 + 1*16.0 + 1*1.008

= 23.976 g/mol

mass of LiOH = 1.31 g

mol of LiOH = (mass)/(molar mass)

= 1.31/23.976

= 0.0546 mol

Balanced chemical equation is:

Cr(NO3)3 + 3 LiOH —> LiNO3 + Cr(OH)3

According to balanced equation

mol of Cr(OH)3 formed = (1/3)* moles of LiOH

= (1/3)*0.0546

= 0.0182 mol

Molar mass of Cr(OH)3,

MM = 1*MM(Cr) + 3*MM(O) + 3*MM(H)

= 1*52.0 + 3*16.0 + 3*1.008

= 103.024 g/mol

mass of Cr(OH)3 = number of mol * molar mass

= 0.0182*103.024

= 1.8763 g

Answer: 1.88 g

7)

Molar mass of BaSO4,

MM = 1*MM(Ba) + 1*MM(S) + 4*MM(O)

= 1*137.3 + 1*32.07 + 4*16.0

= 233.37 g/mol

mass of BaSO4 = 0.89 g

mol of BaSO4 = (mass)/(molar mass)

= 0.89/233.37

= 3.8*10^-3 mol

According to balanced equation

mol of Na2SO4 reacted = moles of BaSO4

= 3.8*10^-3 mol

Answer: 3.8*10^-3 mol

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