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?

Question

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?

Peyton Adams · Accepted Answer

Approx. #13# #mL# of water.

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.

Cameron Andrews · Answer

You will need #"13 g H"_2"O"#.

Balanced Equation #"CaC"_2("s")+"2H"_2"O(l)"##rarr##"C"_2"H"_2("g")+"Ca(OH)"_2("aq")"# To respond to this query, we will perform the conversions listed below. #"mass CaC"_2##rarr##"mol CaC"_2##rarr##"mol H"_2"O"##rarr##"mass H"_2"O"# To do this, we must first find the molar mass of each compound. We can do this by calculation or by consulting a reliable source, such as The PubChem Project. Molar Densities #"CaCl"_2":# #"64.0994 g/mol"# https://tutor.hix.ai #"H"_2"O":# #"18.01528 g/mol"# https://tutor.hix.ai The mole ratio between the two compounds is also required. 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"#. Resolution 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"# The steps can all be combined 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"# Rounding errors are minimized because you can perform the computations all at once and round at the end.

Jose Ayala · Answer

undefined 33 g of H₂O.

In the reaction #CaC_2(s) + 2H_2O(l) -&gt; 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?

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?