Chemistry College

2.56 g of hydrogen reacts completely with 20.32 g of oxygen
to form X g of water. X = g

Answers

Answer 1

Answer:

Mass of water produced is 22.86 g.

Explanation:

Given data:

Mass of hydrogen = 2.56 g

Mass of oxygen = 20.32 g

Mass of water = ?

Solution:

Chemical equation:

2H₂ + O₂   →  2H₂O

Number of moles of oxygen:

Number of moles = mass/ molar mass

Number of moles = 20.32 g/ 32 g/mol

Number of moles = 0.635 mol

Number of moles of hydrogen:

Number of moles = mass/ molar mass

Number of moles = 2.56 g/ 2 g/mol

Number of moles = 1.28 mol

Now we will compare the moles of water with oxygen and hydrogen.

                    O₂            :            H₂O

                     1              :             2

                  0.635        ;            2×0.635 =  1.27

                   H₂             :              H₂O

                    2              :              2

                 1.28            :           1.28

The number of  moles of water produced by oxygen are less thus it will be limiting reactant.

Mass of water produced:

Mass = number of moles × molar mass

Mass = 1.27 × 18 g/mol

Mass = 22.86 g

 

Answer 2

Answer:

22.88

Explanation:

correct have a g'day mate

Related Questions

what makes an element reactive

Answers

Answer:

The number of electrons in the outermost shell of an atom determines its reactivity.

Explanation:

Answer:

The number of electrons in the outermost shell of an atom determines its reactivity. Noble gases have low reactivity because they have full electron shells. Halogens are highly reactive because they readily gain an electron to fill their outermost shell.

Explanation:

* 20 POINTS

A ________ is made of two or more elements chemically combined.


a.

alloy


b.

mixture


c.

element


d.

compound

Answers

Answer:

Compound is the answer

What is the volume of 12.0 M NaOH needed to prepare 5.00 L of a 2.50 M NaOH solution? What volume of water (solvent) is required to make this dilution? Please show work

Answers

Answer:

Approximately [tex]1.04\; \rm L[/tex] of the [tex]12.0\; \rm M[/tex] [tex]\rm NaOH[/tex] solution is needed. Approximately another [tex]3.96\; \rm L[/tex] of water will also be required.

Assumption: the volume of these [tex]\rm NaOH[/tex] solutions does not depend on the quantity of [tex]\rm NaOH\![/tex] (the solute) that each of them contain.

Explanation:

Calculate the number of moles of [tex]\rm NaOH[/tex] formula units in that [tex]5.00\; \rm L[/tex] of [tex]2.50\; \rm M[/tex] [tex]\rm NaOH\![/tex] solution:

[tex]n(\mathrm{NaOH}) = c\cdot V = 5.00\; \rm L \times 2.50\; \rm mol \cdot L^{-1} = 12.5\; \rm mol[/tex].

Calculate the volume of a [tex]12.0\; \rm M[/tex] [tex]\rm NaOH[/tex] with that many [tex]\rm NaOH\![/tex] formula units:

[tex]\displaystyle V = \frac{n}{c} = \frac{12.5\;\rm mol}{12.0\; \rm mol \cdot L^{-1}}\approx 1.04\;\rm L[/tex].

That should be the volume of the [tex]12.0\; \rm M[/tex] [tex]\rm NaOH[/tex] solution needed to prepare that [tex]5.00\; \rm L[/tex] of [tex]2.50\; \rm M[/tex] [tex]\rm NaOH\![/tex] solution. However, [tex]1.04\; \rm L[/tex] corresponds to only about one-fifth the volume of a [tex]5.00\; \rm L\![/tex] solution. The difference in volume should be filled with pure water:

[tex]\begin{aligned}V(\text{water}) &\approx 5.00\; \rm L - 1.04\; \rm L = 3.96\; \rm L\end{aligned}[/tex].

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