If 3.70 mol calcium carbide (CaC2) reacts with an excess of water, how many moles of acetylene (C2H2) will be produced?
To determine the number of moles of acetylene (C2H2) produced, we need to find the mole ratio between calcium carbide (CaC2) and acetylene (C2H2) from the balanced chemical equation.
The balanced chemical equation for the reaction between calcium carbide (CaC2) and water (H2O) to produce acetylene (C2H2) is:
CaC2 + 2H2O -> Ca(OH)2 + C2H2
From the balanced equation, we can see that one mole of calcium carbide (CaC2) reacts with two moles of water (H2O) to produce one mole of acetylene (C2H2).
Therefore, the mole ratio between CaC2 and C2H2 is 1:1.
Given that we have 3.70 mol of CaC2, we can conclude that:
3.70 mol of CaC2 → 3.70 mol of C2H2
To find the number of moles of acetylene (C2H2) produced when 3.70 mol of calcium carbide (CaC2) reacts with an excess of water, we need to use the balanced chemical equation for the reaction.
The balanced chemical equation for the reaction between calcium carbide and water is as follows:
CaC2 + 2H2O → C2H2 + Ca(OH)2
This equation tells us that 1 mol of calcium carbide (CaC2) reacts to produce 1 mol of acetylene (C2H2).
Therefore, if 3.70 mol of calcium carbide (CaC2) is used, we can conclude that 3.70 mol of acetylene (C2H2) will be produced.
So the answer is 3.70 moles of acetylene (C2H2) will be produced.
Write the equation and balance it.
CaC2 + 2H2O ==> C2H2 + Ca(OH)2
Now use the coefficients to convert mols CaC2 to mols of anything in the equation.
3.70 mol CaC2 x (1 mol C2H2/1 mol CaC2) = ? mol C2H2