I also need help on questions 3,7 and 8 too. These are the last questions. Thank you :)

3) How many molecules are in 23 moles of Sodium?

7) How many grams of HCl will be produced with 16.2 grams of H_{2} and excess Cl_{2}?

H_{2} + Cl_{2} -> 2HCl

8) How many Liters are there in 46 g of Cl_{2} gas?

sii1lver Feb 16, 2018

#1**0 **

3) One mole = 6.0221415 x 10^23 molecules

23 x [ 6.0221415 x 10^23] =1.385092545 x 10^24 - molecules in 23 moles of Sodium.

8)

Convert from grams into liters:

46 g of Cl_2 (chlorine)

Convert from grams to cubic centimeters using the density 0.003214 g/cm^3 (at STP) of Cl_2:

46 g (1/(0.003214 g/cm^3)) = 14310 cm^3

Convert to milliliters using the relation 1 cm^3 = 1 mL:

14310 cm^3 ((1 mL)/(1 cm^3)) = 14310 mL

Convert to liters using the relation 1000 mL = 1 L:

14310 mL ((1 L)/(1000 mL)) = **14.31 L**

**7)** Sorry, I don't know this one!.

Guest Feb 16, 2018

#3**+1 **

7) Ok, I will take at stab at it, but you will have to have it checked by some expert!.

Since the molar mass of H_2 =2.016 g/mole, then 16.2 grams =16.2/2.016 =8.04 moles of H_2.

Since the reaction will require the same number of CL_2 moles, or 8.04 moles, it therefore will rquire:

8.04 x 70.9 =~570 grams of CL_2, because 1 mole of CL_2 =70.9 g/mole.

Therefore: 16.2 + 570 =586.2 grams of 2HCl

Guest Feb 16, 2018

#5**+2 **

This method is somewhat different than the guest's because it does not need to consider the amount of Cl_{2} reacted. The general conversion strategy for this particular procedure goes as follows. The table displays the general layout, and all unknowns are marked with question marks:

\(\text{grams}\rightarrow\text{moles}\rightarrow\text{moles}\rightarrow\text{grams}\)

\(16.2g\text{H}_2\) | \(1\text{molH}_2\) | \(?\text{molHCl}\) | \(?g\text{HCl}\) |

\(?g\text{H}_2\) | \(?\text{molH}_2\) | \(1\text{molHCl}\) |

Let's travel through this table column by column. We start with the given information, 16.2gH_{2}, and the eventual goal is to perform a series of conversions. The first column of the table is already finished.

The second column asks the following question: How many grams of H_{2 }are in one mole of H_{2}? In order to answer this question, we have to reference the indispensable periodic table of the elements. I generally use https://www.ptable.com/

as an electronic version of the table.

The atomic mass of an element is also the molar mass represented in grams, so H has a molar mass of 1.008g. However, realize that we are finding the molar mass of H_{2}. The subscript indicates that there are two hydrogen molecules, so double the original molar mass, 1.008g, to obtain the molar mass of H_{2}. \(1\text{molH}_2=1.008g*2=2.016g\).\(\)

The third column is quite a simple step, actually. It compares the molar ratio of the two molecules in question, HCl and H_{2}, in this case. Determining this information requires some basic knowledge of a balanced equation. In the given chemical reaction, it is possible to perceive it in the following sense: One molecule of H_{2} reacts and yields two molecules of HCl. The number of molecules contained in a mole equals Avagadro's constant, or \(1\text{mol}=6.02*10^{23}\text{ molecules}\). If you continue this logic, the original balanced equation indicates \(1\text{molH}_2=2\text{molHCl}\).

The procedure for the fourth column is identical to the procedure for the second column. How many grams of HCl equals one mole of HCl? Because HCl is a compound, the combined mass of the elements equals its molar mass. As aforementioned, H has a mass of 1.008g per mole. Cl, according to the trusty periodic table, has a mass of 35.45g per mole. Therefore, \(1\text{molHCl}=1.008g+35.45g=36.458g\).

After all this work, we have finally determined all the missing values in the original conversion that I suggested earlier. The table now looks complete.

\(16.2g\text{H}_2\) | \(1\text{molH}_2\) | \(2\text{molHCl}\) | \(36.458g\text{HCl}\) |

\(2.016g\text{H}_2\) | \(1\text{molH}_2\) | \(1\text{molHCl}\) |

This table is really a fancy representation of three ratios.

\(16.2g\text{H}_2* \frac{1\text{molH}_2}{2.016g\text{H}_2}* \frac{2\text{molHCl}}{1\text{molH}_2}* \frac{36.458g\text{HCl}}{1\text{molHCl}}\) | First and foremost, let's cancel out all the common units. Doing this shows that the only unit remaining is the desired unit. |

\(16.2* \frac{1}{2.016}* \frac{2}{1}* \frac{36.458g\text{HCl}}{1}\) | Now it is a matter of simplifying. When I input this entire expression into the calculator, I get an answer close to what the guest got. |

\(586.0g\text{HCl}\) | |

I default to this method because it removes the need to approximate halfway through the calculation.

TheXSquaredFactor Feb 17, 2018

#6**0 **

I am willing to share another method for the eighth question

#8)

The method I default to uses the following conversion. I will create another table:

\(46g\text{Cl}_2\) | \(1\text{molCl}_2\) | \(22.4L\text{Cl}_2\text{@STP}\) |

\(?g\text{Cl}_2\) | \(1\text{molCl}_2\) |

There is only one unknown here! We only need to find the number of grams that equals the number of moles of Cl_{2}. There are 35.45 grams of Cl per mole, according to the trusty periodic table. Cl_{2} has double the number of molecules as Cl, so \(1\text{molCl}_2=35.45g*2=70.9g\). In case you are unaware, "STP" is shorthand for standard temperature and pressure.

\(46g\text{Cl}_2*\frac{1\text{molCl}_2}{70.9g\text{Cl}_2}*\frac{22.4L\text{Cl}_2\text{@STP}}{1\text{molCl}_2}\) | Cancel out all the units. |

\(46*\frac{1}{70.9}*\frac{22.4L\text{Cl}_2\text{@STP}}{1}\) | The rest is a calculator's job. Yet again, the guest answer and my answer are close. |

\(14.53L\text{CL}_2\) |

TheXSquaredFactor Feb 17, 2018