Answer:
2.67g of NH4Cl are required
Explanation:
Molarity is an unit of concentration in chemistry defined as the moles of solute (In this case, NH4Cl), per liter of solution.
To prepare 100mL of a 0.50M are required:
100mL = 0.100L * (0.50 moles / L) = 0.0500 moles NH4Cl
As molar mass of NH4Cl is 53.491g/mol:
0.0500 moles * (53.491g / mol) =
<h3>2.67g of NH4Cl are required</h3>
Answer:
1) Write the balanced equation:
2C2H6 + 7O2 ---> 4CO2 + 6H2O
2) Determine limiting reagent:
C2H6 ⇒ 13.8 g / 30.0694 g/mol = 0.45894 mol
O2 ⇒ 45.8 g / 31.9988 g/mol = 1.4313 mol
C2H6 ⇒ 0.45894 / 2 = 0.22947
O2 ⇒ 1.4313 / 7 = 0.20447
Oxygen is limiting.
3) Determine theoretical yield of water:
The oxygen : water molar ratio is 7 : 6
7 is to 6 as 1.4313 mol is to x
x = 1.2268286 mol of water
4) Convert moles of water to grams:
1.2268286 mol times 18.015 g/mol = 22.1 g (to three sig figs)
Solution to (b):
14.2 g / 22.1 g = 64.2%
Explanation:
Taking into account the reaction stoichiometry, 340.0 moles of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas
<h3>Reaction stoichiometry</h3>
In first place, the balanced reaction is:
CO₂ + 4 H₄ → CH₄ + 2 H₂O
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- CO₂: 1 mole
- H₄: 4 moles
- CH₄: 1 mole
- H₂O: 2 moles
<h3>Moles of CH₄ formed</h3>
The following rule of three can be applied: if by reaction stoichiometry 1 mole of CO₂ form 4 moles of CH₄, 85.1 moles of CO₂ form how many moles of CH₄?
<u><em>moles of CH₄= 340.4 moles</em></u>
Then, 340.0 moles of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas
Learn more about the reaction stoichiometry:
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Answer;
Mass in gram/molecular mass=no.of molecule of Cl2 gas/Avogadro’s number(NA) - - - - - - - - -(1)
Here,
Molecular mass of cl2=35.5*2=71
No.of cl2 molecule=5*10^22
Avogadro’s number(NA)=6.032*10^23
Now,substituting these all value in equation one,we get,
Mass in gram =(no.of molecule)/NA)*molecular mass
Or,Mass in gram =(5*10^22/6.023*10^23)*71
Or,Mass in gram = 5.89gm.Ans
