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For the balanced equation shown below, if 40.7 grams of Al were reacted with 86.6 grams of HCl, how many grams of AlCl3 would be produced?
2Al+6HCl=>2AlCl3+3H2
1.Find limiting reagent
2Al + 6HCl
40.7g 86.6g
27g/mol 36.5g/mol
1.50=mol 2.37=mol
x 6 x 2
9.00 4.74
The limiting reagent is 6HCl
2.Figure out the molar mass of 2AlCl3
Alx1
Clx3
27+106.5=133.5
3.Compare the limiting reagent (6HCl) with 2AlCl3 and find ratio
6HCl 2AlCl3
6:2
4.Divide ratio of 6HCl and 2AlCl3
1:.33
5.Take ratio (1:.33) and multiply with the limiting reagent MOL and molar mass of 2AlCl3
1:.33x2.37x133.5
=104.410
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2Al+6HCl=>2AlCl3+3H2
1.Find limiting reagent
2Al + 6HCl
40.7g 86.6g
27g/mol 36.5g/mol
1.50=mol 2.37=mol
x 6 x 2
9.00 4.74
The limiting reagent is 6HCl
2.Figure out the molar mass of 2AlCl3
Alx1
Clx3
27+106.5=133.5
3.Compare the limiting reagent (6HCl) with 2AlCl3 and find ratio
6HCl 2AlCl3
6:2
4.Divide ratio of 6HCl and 2AlCl3
1:.33
5.Take ratio (1:.33) and multiply with the limiting reagent MOL and molar mass of 2AlCl3
1:.33x2.37x133.5
=104.410
Still don't understand?
Click Here to look at step by step examples
Click Here to watch a video on how to find the limiting reactant