In the reaction #CaC_2(s) + 2H_2O(l) -> C_2H_2(g) + Ca(OH)_2(aq)#, if 23 g of #CaC_2# are consumed in this reaction, how much #H_2O# is needed?

Moles of calcium carbide: #(23*g)/(64.10*g*mol^-1)# #=# #0.359# #mol#.

Given the equation, #"2 equiv"# of water are required:

#0.359xx2*mol# water, #=# #0.718# #mol# water.

#"Equivalent mass"# #=# #0.718*molxx18.01*g*mol^-1# #=# #"approx. 12.9 g"# water.

Balanced Equation
#"CaC"_2("s")+"2H"_2"O(l)"##rarr##"C"_2"H"_2("g")+"Ca(OH)"_2("aq")"#

We will make the following conversions to answer this question.

#"mass CaC"_2##rarr##"mol CaC"_2##rarr##"mol H"_2"O"##rarr##"mass H"_2"O"#

In order to do this, we need to determine the molar mass of each compound. We can calculate it or look it up on a reputable source. I like to use The PubChem Project.

Molar Masses

#"CaCl"_2":# #"64.0994 g/mol"#
https://pubchem.ncbi.nlm.nih.gov/compound/6352#section=Top

#"H"_2"O":# #"18.01528 g/mol"#
https://pubchem.ncbi.nlm.nih.gov/compound/962#section=Top

We also need the mole ratio between the two compounds.

From the balanced equation we can see that the mole ratio between #"CaC"_2"# and #"H"_2"O"# is #"1 mol CaC"_2: "2 mol H"_2"O"#.

Solution

Convert #"mass CaC"_2##rarr##"mol CaC"_2# by dividing the given mass by its molar mass.

#23cancel"g CaC"_2xx(1"mol CaC"_2)/(64.0994cancel"g CaCl"_2)="0.358817711242 mol CaC"_2"# (I will round to two significant figures at the end. Meanwhile I'm keeping the calculator results until the end.

Convert #"mol CaC"_2"##rarr##"mol H"_2"O"# by multiplying the mol #"CaC"_2"# by the mole ratio with water in the numerator.

#0.358817711242cancel"mol CaC"_2xx(2"mol H"_2"O")/(1cancel"mol CaC"_2)="0.717635422484 mol H"_2"O"#

Convert #"mol H"_2"O"##rarr##"mass H"_2"O"# by multiplying the mol #"H"_2"O"# by its molar mass.

#0.717635422484cancel"mol H"_2"O"xx(18.01528"g H"_2"O")/(1cancel"mol H"_2"O")=13"g H"_2"O"#

We can put all of the steps together like this:

#23cancel"g CaC"_2xx(1"mol CaC"_2)/(64.0994cancel"g CaC"_2)xx(2"mol H"_2"O")/(1cancel"mol CaC"_2)xx(18.01528"g H"_2"O")/(1cancel"mol H"_2"O")="13 g H"_2"O"#

This way you can do the calculations all at once and round at the end, which reduces rounding errors.