Limiting Reactant problem #2. 10 points for first (Correct) answer!!!?

Question

Carbon reacts with steam (H2O) under certain conditions to produce hydrogen and carbon monoxide.

a) If 2.4 moles of carbon reacts with 3.10 moles of steam, identify the limiting reactant.

b) How many moles of each product are formed?

c) How many grams of each product are formed?

Answer 1

C + H2O ===> CO + H2

a) C is the limiting reactant, because the equation dictates a 1:1 ratio, and C is present in lesser amount.

b) 2.4 moles H2 + 2.4 moles CO

c) Atomic weights: C=12 H=1 O=16 CO=28 H2=2

2.4molH2 x 2gH2/1molH2 = 4.8g H2

2.4molCO x 28gCO/1molCO = 67.2g CO

Answer 2

the formula will be
C + H2O ---> H2 + CO

Since it is both 1 to 1 mole ratio for both C and H2O to produce CO... which ever reactant with less mole is the limiting reactant... in this case, Carbon

only 2.4 mole of CO can be formed... and 4.8 mole of H can be formed

you convert the 2.4 mole of CO into gram by mulitplying 2.4 by 28 (12---C+16---O), you get 67.2 gram... for 4.8 mole H, you multiply 4.8 by 1.01, you get 4.848

Answer 3

2C+3H2O---> CO2 + CO + 3H2

a.(2.4 mol C) times( 3 mol H2O / 2 mol C)=3.6 mol of H2O need for 2.4 mol of C. so limiting reactant is H2O

b.1.03 mol of carbon dioxide, 1.03 mol of carbon monoxide
3.10 mole of H2 gas.

(3.10 mol H2O) times ( 1 mol CO2 / 3 mol of H2O)=1.03 mol CO2

c. well just convert the moles to grams through molar mass ,,\/(^.~)