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?

Answer 1

Peyton Adams

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.

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Answer 2

Cameron Andrews

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://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

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.

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Answer 3

Jose Ayala

33 g of H₂O.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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?