Given the general chemical equation 3A + 2B → C + 5D,
1 answer:
Answer :
(a) The number of moles of D produced can be, 6.67 moles.
(b) The volume of D prepared can be, 24.5 L
Explanation :
The given chemical reaction is:
Part (a) :
From the balanced chemical reaction, we conclude that:
As, 3 moles of A react to give 5 moles of D
So, 4 moles of A react to give moles of D
Thus, the number of moles of D produced can be, 6.67 moles.
Part (b) :
As we know that 1 moles of substance occupies 22.4 L volume of gas.
As, volume of B gives
volume of D
As, 9.8 L volume of B gives volume of D
Thus, the volume of D prepared can be, 24.5 L
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Here we have to choose the right option which tells the moles of CaCl₂ will react with 6.2 moles of AgNO₃ in the reaction
2AgNO₃ + CaCl₂→ 2AgCl + Ca(NO₃)₂
6.2 moles of silver nitrate (AgNO₃) will react with B. 3.1 moles of calcium chloride (CaCl₂).
From the reaction: 2AgNO₃ + CaCl₂→ 2AgCl + Ca(NO₃)₂
Thus 2 moles of AgNO₃ reacts with 1 mole of CaCl₂
Henceforth, 6.2 moles of AgNO₃ reacts with = 3.1 moles of CaCl₂.
1 mole of CaCl₂ reacts with 2 moles of AgNO₃. Thus-
A. 2.2 moles of CaCl₂ will react with 2.2×2 = 4.4 moles of AgNO₃.
C. 6.2 moles of CaCl₂ will reacts with 6.2×2 = 12.4 moles of AgNO₃.
D. 12.4 moles of CaCl₂ will reacts with 12.4 × 2 = 24.8 moles of AgNO₃
Thus the right answer is 6.2 moles of AgNO₃ will react with 3.1 moles of CaCl₂.
N2O4(l) + 2 N2H4(l) → 3 N2(g) + 4 H2O(g)
1) to calculate the limiting reactant you need to pass grams to moles.
<span> moles is calculated by dividing mass by molar mass
</span>
mass of N2O4: 50.0 g
molar mass of <span>N2O4 = 92.02 g/mol
</span><span>molar mass of N2H4 = 32.05 g/mol.
</span>mass of N2H4:45.0 g
moles N2O4=50.0/92.02 g/mol= 0,54 mol of N2O4
moles N2H4= 45/32.05 g/mol= 1,40 mol of <span><span>N2H4
</span> 2)</span>
By looking at the balanced equation, you can see that 1 mol of N2O4 needs 2 moles of N2H4 to fully react . So to react 0,54 moles of N2O4, you need 2x0,54 moles of <span>N2H4 moles
</span><span>N2H4 needed = 1,08 moles.
You have more that 1,08 moles </span><span>N2H4, so this means the limiting reagent is not N2H4, it's </span>N2O4. The molecule that has molecules that are left is never the limiting reactant.
3) 1 mol of N2O4 reacting, will produce 3 mol of N2 (look at the equation)
There are 0,54 mol of N2O4 available to react, so how many moles will produce of N2?
1 mol N2O4------------3 mol of N2
0,54 mol N2O4--------x
x=1,62 mol of N2
4) the only thing left to do is convert the moles obtained, to grams.
We use the same formula as before, moles equal to mass divided by molar mass.
(molar mass of N2= 28)
1,62 mol of N2= mass/ 28
mass of N2= 45,36 grams
1) to calculate the limiting reactant you need to pass grams to moles.
<span> moles is calculated by dividing mass by molar mass
</span>
mass of N2O4: 50.0 g
molar mass of <span>N2O4 = 92.02 g/mol
</span><span>molar mass of N2H4 = 32.05 g/mol.
</span>mass of N2H4:45.0 g
moles N2O4=50.0/92.02 g/mol= 0,54 mol of N2O4
moles N2H4= 45/32.05 g/mol= 1,40 mol of <span><span>N2H4
</span> 2)</span>
By looking at the balanced equation, you can see that 1 mol of N2O4 needs 2 moles of N2H4 to fully react . So to react 0,54 moles of N2O4, you need 2x0,54 moles of <span>N2H4 moles
</span><span>N2H4 needed = 1,08 moles.
You have more that 1,08 moles </span><span>N2H4, so this means the limiting reagent is not N2H4, it's </span>N2O4. The molecule that has molecules that are left is never the limiting reactant.
3) 1 mol of N2O4 reacting, will produce 3 mol of N2 (look at the equation)
There are 0,54 mol of N2O4 available to react, so how many moles will produce of N2?
1 mol N2O4------------3 mol of N2
0,54 mol N2O4--------x
x=1,62 mol of N2
4) the only thing left to do is convert the moles obtained, to grams.
We use the same formula as before, moles equal to mass divided by molar mass.
1,62 mol of N2= mass/ 28
mass of N2= 45,36 grams