The reaction involved in this problem is called the combustion reaction where a hydrocarbon reacts with oxygen to product carbon dioxide and water. The reaction of C2H5OH would be as follows:
C2H5OH + 3O2 = 2CO2 + 3H2O
To determine the number of molecules of CO2 that is formed, we need to determine the number of moles produced from the initial amount of C2H5OH and the relation from the reaction. Then we multiply avogadros number which is equal to 6.022x10^23 molecules per mole.
2.00 g C2H5OH ( 1 mol C2H5OH / 46.08 g C2H5OH ) ( 2 mol CO2 / 1 mol C2H5OH ) = 0.0868 mol CO2
0.0868 mol CO2 ( 6.022x10^23 molecules / mol ) = 5.23x10^22 molecules CO2
NaCl is table salt so you need 4Na and 4 Cl, So Na₄Cl₄
Hi!
• For this you want to convert moles to grams
• To do this you would simply multiple your moles times the atomic mass of potassium
3.25 x 39.098= 127.0685
• So your answer for this should be about 127.07 depending on how you need to round!
Answer:
The value of the mole is equal to the number of atoms in exactly 12 grams of pure carbon-12. 12.00 g C-12 = 1 mol C-12 atoms = 6.022 × 1023 atoms • The number of particles in 1 mole is called Avogadro's Number (6.0221421 x 1023).
Explanation:
Answer:
18.7887 g of NaCl
Explanation:
<em>The question reads - How many grams of NaCl will be produced from 18.0 g of Na and 23.0 g of Cl2?</em>
Let us start by writing out the balanced equation of the reaction:
Na + Cl2 ---> NaCl2
1 mole, each of Na and Cl2 is required to produce 1 mole of NaCl.
mole = mass/molar mass
Therefore
18 g of Na = 18/23 = 0.7826 mole
23 g of Cl2 = 23/71 = 0.3239 mole
In this case, the Na is in excess and the Cl2 becomes the limiting reagent. Hence
0.3239 mole of Cl2 will react with 0.3239 mole of Na to yield 0.3239 mole of NaCl.
mass of 0.3239 mole NaCl = 0.3239 x 58 = 18.7887 g
<u>Hence, 18.7887 grams of NaCl will be produced from 18.0 g of Na and 23.0 g of Cl2.</u>