If 4.70 mol of calcium carbide (CaC2) reacts with an excess of water, how many moles of acetylene (C2H2), a gas used in welding, will be produced?
CaC2(s) + 2 H2O(l) Ca(OH)2(aq) + C2H2(g)
mol
To find out how many moles of acetylene (C2H2) will be produced when 4.70 mol of calcium carbide (CaC2) reacts with an excess of water, we need to use the stoichiometry of the balanced chemical equation.
First, let's identify the stoichiometric ratio between CaC2 and C2H2. According to the balanced equation:
1 mole of CaC2 reacts to produce 1 mole of C2H2.
Since the stoichiometry is 1:1 between CaC2 and C2H2, we can conclude that the number of moles of C2H2 produced will be the same as the moles of CaC2 used in the reaction.
Therefore, 4.70 moles of acetylene (C2H2) will be produced when 4.70 moles of calcium carbide (CaC2) reacts with an excess of water.
You need to find the arrow key and use it.
CaC2(s) + 2H2O(l)=> Ca(OH)2(aq) + C2H2(g)
Use the coefficients to convert anything to any other reactant or product.
4.7 mol CaC2 x (1 mol C2H2/1 mol CaC2) = 4.7 x 1/1 = 4.7 mol C2H2.
Note that the units you don't want cancel (that's mol CaC2) and leaves the unit you want to convert to (C2H2).
4.7 mols CaC2 will required 2*4.7 mols H2O and will produce 1 mol Ca(OH)2.