a For the following unbalanced reaction suppose exactly 5.30 g of each reactant

Question

a For the following unbalanced reaction suppose exactly 5.30 g of each reactant is taken. Determine which reactant is limiting, and also determine what mass of the excess reagent will remain after the limiting reactant is consumed. Na B 07 (8) + H SO (aq) + H 0(1) HBO3(s) + Na SO. (aq) Limiting reactant: Mass of excess reagent 1 = Mass of excess reagent 2 = Submit b For the following unbalanced reaction, suppose exactly 6.55 g of each reactant is taken. Determine which reactant is limiting, and also determine what mass of the excess reagent will remain after the limiting reactant is consumed. CaCz(s) + H2O(l) ? Ca(OH)2 (8) + C2H2 (9) Limiting reactant Mass of excess reagent : C For the following unbalanced reaction suppose exactly 6.30 g of each reactant is taken. Determine which reactant is limiting, and also determine what mass of the excess reagent will remain after the limiting reactant is consumed. NaCl(s) + H, SO (1) ? HCI(9) + Nazso(s) Limiting reactant: Mass of excess reagent = Previous Next

Explanation / Answer

1) Na2B4O7 + H2SO4 + 5H2O ------> 4H3BO3 + Na2SO4

No. of moles of Na2B4O7 = 5.30/201.219 = 0.02634

No of moles of H2SO4 = 5.30/98.079 = 0.0540

No of moles of H2O = 5.30/18.01 = 0.2942

As the molar ratio of Na2B4O7 : H2SO4 : H2O = 1:1:5 in the above reaction

1 mole of Na2B4O7 ---- 5 moles of H2O  

0.02634 moles of Na2B4O7 ------ 5x 0.02634 = 0.1317

the no. of moles of water are also in excess it is excess reagent 1 i; no of moles remained after limiting reactant is consumed = 0.2942-0.02634 = 0.26786

its mass in gms = no. of moles x mol wt = 0.26786x18.01 = 4.84 gms - excess reagent 1

Excess reagent 2 is H2SO4 its no. of moles in excess = 0.0540- 0.02634 = 0.02766

Its mass = no. of moles x mol wt = 0.02766 x 98.079 = 2.712 gms = mass of excess reagent 2

So the number of moles Na2B4O7 is less in number so it is limiting reagent

2) CaC2 + 2H2O ---> Ca(OH)2 + C2H2

molar ratio of calcium carbide :water = 1:2

No. of moles of CaC2 = 6.55/64.099 = 0.10218

No. of moles of water = 6.55/18.01 = 0.3637

1 mole of CaC2 react to 2 mole of water

0.10218 mole of CaC2 ---- 0.10218 X 2 = 0.20436

So the number of moles CaC2 is low in number than water so it is limiting reagent

no.of moles of excess reagent will remain after the limiting reactant = 0.3637-0.10218 = 0.26152

Mass of excess reagent = no. of moles x mol wt = 0.26152 x 18.01 = 4.709 gms

3) 2NaCl + H2SO4 ---> 2HCl + Na2SO4

No. of moles of NaCl = 6.30/58.4 = 0.1078

No. of mole of H2SO4 = 6.30/98.079 = 0.0642

molar ratio of NaCl : H2SO4 = 2:1

2 moles of NaCl --- 1 mole of H2SO4

0.1078 moles of NaCl -- ? 0.0539 moles of H2SO4

As the no of moles of H2SO4 is more in number so limiting reagent is NaCl

no. of moles of excess reagent will remain after limiting reactant is consumed = 0.0642-0.0539 = 0.0103

mass of excess reagent = 0.0103 x 98.079 = 1.010 gms

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