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For the balanced equation shown below, if 32.8 grams of FeS were reacted with 21.1 grams of HCl, how many grams of H2S would be produced?
FeS+2HCl=>H2S+FeCl2
1.Find limiting reagent
FeS + 2HCl
32.8g 21.1g
91g/mol 36.5g/mol
.360=mol .578=mol
x 2 x 1
.720 .578
The limiting reagent is 2HCl
2.Figure out the molar mass of H2S
Hx2
Sx1
2+32=34
3.Compare the limiting reagent (2HCl) with H2S and find ratio
2HCl 1H2S
2:1
4.Divide ratio of 2HCl and H2S
2:1
2 2
1:.5
5.Take ratio (1:.5) and multiply with the limiting reagent MOL and molar mass of H2S
1:.5x.578x34
=9.826
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FeS+2HCl=>H2S+FeCl2
1.Find limiting reagent
FeS + 2HCl
32.8g 21.1g
91g/mol 36.5g/mol
.360=mol .578=mol
x 2 x 1
.720 .578
The limiting reagent is 2HCl
2.Figure out the molar mass of H2S
Hx2
Sx1
2+32=34
3.Compare the limiting reagent (2HCl) with H2S and find ratio
2HCl 1H2S
2:1
4.Divide ratio of 2HCl and H2S
2:1
2 2
1:.5
5.Take ratio (1:.5) and multiply with the limiting reagent MOL and molar mass of H2S
1:.5x.578x34
=9.826
Still confused? Click Here to go to the sample problems!