Answer:
The limiting reactant is the 6.279 g of ![NiCl_2](https://tex.z-dn.net/?f=NiCl_2)
Explanation:
We have to start with the <u>reaction</u> between sodium carbonate (
) and the Nickel (II) Chloride (
), so:
![Na_2CO_3~+~NiCl_2-->~NiCO_3~+~NaCl](https://tex.z-dn.net/?f=Na_2CO_3~%2B~NiCl_2--%3E~NiCO_3~%2B~NaCl)
We will have a <u>double replacement reaction</u>. Now we have to <u>balance</u> the reaction, so:
![Na_2CO_3~+~NiCl_2-->~NiCO_3~+~2NaCl](https://tex.z-dn.net/?f=Na_2CO_3~%2B~NiCl_2--%3E~NiCO_3~%2B~2NaCl)
The next step is the <u>calculation of the moles for each reactive</u>. For
we have use the <u>molarity equation</u>:
![M~=~\frac{mol}{L}](https://tex.z-dn.net/?f=M~%3D~%5Cfrac%7Bmol%7D%7BL%7D)
![0.1010~M~=\frac{mol}{0.5~L}](https://tex.z-dn.net/?f=0.1010~M~%3D%5Cfrac%7Bmol%7D%7B0.5~L%7D)
![mol~=~0.1010*0.5=~0.0505~mol~of~Na_2CO_3](https://tex.z-dn.net/?f=mol~%3D~0.1010%2A0.5%3D~0.0505~mol~of~Na_2CO_3)
For the calculation of moles of
we have to use the <u>molar mass</u> of the compound (129.59 g/mol):
![6.279~g~NiCl_2\frac{1~mol~NiCl_2}{129.59~g~NiCl_2}=~0.0484~mol~NiCl_2](https://tex.z-dn.net/?f=6.279~g~NiCl_2%5Cfrac%7B1~mol~NiCl_2%7D%7B129.59~g~NiCl_2%7D%3D~0.0484~mol~NiCl_2)
The next step is the division of each mole value by the <u>coefficient of each reactive</u> of the balance reaction. In this case <u>we have "1" for each reactive</u>, so:
![\frac{0.0484}{1}=0.0484](https://tex.z-dn.net/?f=%5Cfrac%7B0.0484%7D%7B1%7D%3D0.0484)
![\frac{0.0505}{1}=0.0505](https://tex.z-dn.net/?f=%5Cfrac%7B0.0505%7D%7B1%7D%3D0.0505)
The final step is to <u>choose the smallest value</u>. In this case is the value that correspond to
. Therefore
is the limiting reactive.
CH4 - no lone pairs
HCl - 3 lone pairs
H2O - 2 lone pairs
NH3 - 1 lone pair (answer)
The first step is to make a balanced chemical equation.
2AgNO3 + CaCl2 ---> 2AgCl + Ca(NO3)2
Molecular Weights:
CaCl2 = 110.98 g/mol
AgNO3 =170.01
AgCl: 143.45 g/mol
Volume:
CaCl2: 30.0mL=0.03L
AgNO3: 15.0mL=0.015 L
Solving for the limiting reactant one needs to get the mols CaCl2 and mols AgNO3:
CaCl2: 0.150M(mol/L) * 0.03L = 0.0045 moles
AgNO3: 0.100M*0.015L = 0.0015 moles
Since the stoichiometric ratio of AgNO3 to CaCl2 is 2:1
0.0015 mols AgNO3 *(1 mol CaCl2/ 2 mols AgNO3) = 0.00075 mols CaCl2
Since the answer is lesser than CaCl2 then the limiting reactant is AgNO3.
To get the mass of AgCl one will do a stoichiometric calculation with respect to the limiting reactant, AgNO3.
0.0015 moles AgNO3 *
Step 1. - Scientists first take a look at their empirical evidence and try to explain it.
Step 2. - Scientists test if their explanation or conclusion is logical.
Step 3. - If their explanation is logical, they search for new test that may help them to support their first idea.
<span>we know that each
element has an unique spectra and it can be used to identify the
element. it shows that the energy levels of the electrons and different colors are the result of different wavelengths.
hope it helps
</span>