Question

Chemistry Honors:
Calculations Involving a Limiting Reactant
For each of the following unbalanced chemical equations, suppose that exactly 5.00 g of each reactant is taken. Determine which reactant is limiting, and calculate what mass of each product is expected (assuming that the limiting reactant is completely consumed).
S(s) + 2H2SO4(aq) --> 3SO2(g) + 2H2O(l)

Answer 1

In limiting reactant problems, ALWAYS conduct calculations in moles.

molar mass S(s) = 1(32.065) = 32.065 grams/mole
5.00 grams of S(s) => 5.00/32.065 = 0.156 moles of S(s)
molar mass H2SO4(aq) = 2(1.00794) + 1(32.065) + 4(15.9994) = 98.079 grams/mole
5.00 grams of H2SO4(aq) => 5.00/98.079 = 0.0510 moles of H2SO4(aq)

According to the balanced chemical reaction, H2SO4(aq) is required at twice the quantity of S(s), so the comparison is:
0.156 moles S(s) vs. 0.102 moles H2SO4(aq)
Since less H2SO4(aq) than S(s) is available, it determines how much products will be produced. This is also called being the limiting reactant.

Given 0.102 moles H2SO4(aq), how much SO2(g) will form?
SO2(g) is formed at 3/2 the rate that H2SO4(aq) is used:
(3/2)(0.102 moles H2SO4(aq)) => 0.153 moles SO2(g)
The molar mass of SO2(g) = 1(32.065) + 2(15.9994) = 64.0638 grams/mole
(0.153)(64.0638) = 9.80 grams SO2(g)

Given 0.102 moles H2SO4(aq), how much H2O(l) will form?
H2O(l) is formed at the same rate that H2SO4(aq) is used:
0.102 moles H2SO4(aq) => 0.102 moles H2O(l)
The molar mass of H2O(l) = 2(1.00794) + 1(15.9994) = 18.015 grams/mole
(0.102)(18.015) = 1.84 grams H2O(l)

Answer 2

can you show me how to limiting reactant with 2 moles of h2so4