Answer is: not enough <span>colorless syrupy liquid.
</span>n(H₂SO₄) = 1,2 mol.
M(H₂SO₄) = 2Ar(H) + Ar(S) + 4Ar(O) · g/mol.
M(H₂SO₄) = 2·1 + 32 + 4·16 · g/mol.
M(H₂SO₄) = 98 g/mol.
m(H₂SO₄) = n(H₂SO₄) · M(H₂SO₄).
m(H₂SO₄) = 1,2 mol · 98 g/mol.
m(H₂SO₄) = 117,6 g needed.
100 g is less that 117,6 g.
Solution :
Molar mass of 
is :
M = 6×12 + 6×1 g
M = 78 g
78 gram of contains 
molecules.
So, 89.5 gram of contains :
Now, from the formula we can see that one molecule of contains 2 hydrogen atom . So, number of hydrogen atom are :
Hence, this is the required solution.
Answer:
116.3 grCO2
Explanation:
1st - we balance the equation so that it finds the same amount of elements of the product side and of the reagent side
C6H6 +15/2 O2⟶ 6CO2 +3 H2O
2nd - we calculate the limiting reagent
39.2gr C6H6*(240grO2/78grC6H6)=120 grO2
we don't have that amount of oxygen so this is the excess reagent and oxygen the limiting reagent
3rd - we use the limiting reagent to calculate the amount of CO2 in grams
105.7grO2*(264grCO2/240grO2)=116.3 grCO2
