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

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6

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 H2 and excess Cl2?

H2 + Cl2 -> 2HCl

8) How many Liters are there in 46 g of Cl2 gas?

off-topic
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!.

Feb 16, 2018
#2
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Thanks!!

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

Feb 16, 2018
#4
+1

Thank you soooo much!!

Feb 17, 2018
#5
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This method is somewhat different than the guest's because it does not need to consider the amount of Cl2 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, 16.2gH2, 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 Hare in one mole of H2? 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 H2. The subscript indicates that there are two hydrogen molecules, so double the original molar mass, 1.008g, to obtain the molar mass of H2$$1\text{molH}_2=1.008g*2=2.016g$$.

The third column is quite a simple step, actually. It compares the molar ratio of the two molecules in question, HCl and H2, 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 H2 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.

Feb 17, 2018
edited by TheXSquaredFactor  Feb 17, 2018
edited by TheXSquaredFactor  Feb 17, 2018
#6
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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 Cl2. There are 35.45 grams of Cl per mole, according to the trusty periodic table. Cl2 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. $$14.53L\text{CL}_2$$
Feb 17, 2018