Question

For each of the following unbalanced reactions, suppose 5.18 g of each reactant is taken. Determine which reactant is limiting, and also determine what mass of the excess reagent(s) will remain after the limiting reactant is consumed. a) CaC2(s) + H2O(l) → Ca(OH)2(s) + C2H2(g) b) NaCl(s) + H2SO4(l) → HCl(g) + Na2SO4(s) c) SiO2(s) + C(s) → Si + CO(g)

Answer 1

a) CaC₂ + 2 H₂O = Ca(OH)₂ + C₂H₂

1 mol of CaC2 requires 2 moles of water. For every 1 mol of water 0.5 moles of CaC2 is required

5.18 g of each reactant is taken

mol of CaC2 = 5.18 / 64 = 0.0808

moles of water = 5.18/18 = 0.287 clearly CaC2 is the limiting reagent

0.0808 moles of CaC2 requires 0.1616 mole of water.

Water remaining = 0.287 - 0.1616 = 0.1254 mol = 2.257 g

b) 2 NaCl + H₂SO₄ = 2 HCl + Na₂SO₄

mole of NaCl = 5.18 / 58.4 = 0.0886

moles of H2SO4 = 5.18 / 98 = 0.0528

NaCl is the limiting reagent

0.0886 mol of NaCl requires 0.0886/2 =0.0443 mol of H2SO4

Remaining H2SO4 = 0.0528 - 0.0443 = 0.0085 mol = 0.0085 x 98 g =0.833 g

c) SiO₂(s) + 2 C(s) = Si + 2 CO(g)

moles of SiO2 = 5.18/60=0.0862

moles of C = 5.18/12 = 0.431

Clearly SiO2 is the limiting reagent

0.0862 moles of SiO2 requires 0.0862 x 2 moles of C = 0.1724

remaining moles of C = 0.431 - 0.1724 = 0.2586mol = 3.1032 g