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1.For the balanced equation shown below, what would be the limiting reagent if 46.3 grams of C3H6O were reacted with 73.2 grams of O2?
C3H6O+4O2=>3CO2+3H2O First find the molar mass.
C = 12
H = 1
O = 16
12 * 3 = 36
1 * 6 = 6
16 * 1 = 16
36 + 6 + 16 = 58 grams/mole
16 * 2 = 32
32 grams/mole
46.3/58 = .798 moles
73.2/32 = 2.2875 moles
1 * .7984
4 2.2875 2.2875 = O2 3.192 = C3H6O
limiting reagent = O2.
2. For the balanced equation shown below, what would be the limiting reagent if 11.5 grams of C6H6O3 were reacted with 6.57 grams of O2?
C6H6O3+3O2=>6CO+3H2O
C = 12
H = 1
O = 16
12 * 6 = 72
1 * 6 = 6
16 * 3 = 48
72 + 6 + 48 = 128 grams/mole
16 * 2 = 3232 grams/mole
11.5/128 = .089 moles
6.57/32 = .205 moles
1 * .0893
3 .205
.205 = O2 .267 = C6H6O3
limiting reagent = O2
3.For the balanced equation shown below, what would be the limiting reagent if 41.9 grams of C2H3OF were reacted with 61.0 grams of O2?
C2H3OF+2O2=>2CO2+H2O+HF
C = 12
H = 1
O = 16
F = 19
12 * 2 = 24
1 * 3 = 3
16 * 1 = 16
19 * 1 = 19
24 + 3 + 16 + 19 = 62 grams/mole
16 * 2 = 32
32 grams/mole
41.9/62 = .675 moles
61/32 = 1.9 moles
1 * .675
2 1.9
1.35 = C2H3OF 1.9 = O2
Limiting reagent = C2H3OF
4.For the balanced equation shown below, what would be the limiting reagent if 26.0 grams of C3H9N were reacted with 46.3 grams of O2?
4C3H9N+25O2=>12CO2+18H2O+4NO2
C = 12
H = 1
N = 14
O = 16
12 * 3 = 36
1 * 9 = 9
14 * 1 = 14
36 + 9 + 14 = 59 grams/mole
16 * 2 = 32
32 grams/ mole
26/59 = .44 moles
46.3/32 = 1.44 moles
4 * .44
25 1.44
11 = C3H9N 5.76 = O2
limiting reagent = O2
5.For the balanced equation shown below, what would be the limiting reagent if 19.2 grams of C7H16 were reacted with 120 grams of O2?
C7H16+11O2=>7CO2+8H2O
C = 12
H = 1
O = 16
12 * 7 = 84
1 * 16 = 16
84 + 16 = 100 grams/mole
16 * 2 = 32 grams/mole
19.2/100 = .192 moles
120/32 = 3.75 moles
1 * .192
11 3.75
2.112 = C7H16 3.75 = O2
limiting reagent = C7H16
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