Answer: Theoretical Yield = 0.2952 g
Percentage Yield = 75.3%
Explanation:
Calculation of limiting reactant:
n-trans-cinnamic acid moles = (142mg/1000) / 148.16 = 9.584*10⁻⁴ mol
pyridium tribromide moles = (412mg/1000) / 319.82= 1.288*10⁻³ mol
- n-trans-cinnamic acid is the limiting reactant
The molar ratio according to the equation mentioned is equals to 1:1
The brominated product moles is also = 9.584*10⁻⁴ mol
Theoretical yield = (9.584*10⁻⁴ mol) * (Mr of brominated product)
= (9.584*10⁻⁴ mol) * (307.97) = 0.2952 g
Percentage Yield is : Actual Yield / Theoretical Yield = 0.2223/0.2952
= 75.3%
C2H6O + O2 ---> C2H4O2 + H2O
using the molar masses:-
24+ 6 + 16 g of C2H6O produces 24 + 4 + 32 g C2H4O2 (theoretical)
46 g produces 60g
60 g C2H4O2 is produced from 46g C2H6O
1g . .................................46/60 g
700g ................................. (46/60) * 700 Theoretically
But as the yield is only 7.5%
the required amount is ((46/60) * 700 ) / 0.075 = 7155.56 g
= 7.156 kg to nearest gram. Answer
<span>choices are:
Zirconium metal plus hydrogen chloride yields zirconium chloride solution and hydrogen gas
Xenon solid plus hydrochloric acid yields xenon chloride and hydrogen gas
Zinc metal plus hydrogen chloride yields zinc dichloride plus hydrogen gas
Zinc metal plus an aqueous solution of hydrochloric acid yields an aqueous solution of zinc chloride plus hydrogen gas
right ans is:
</span>
Zinc metal plus an aqueous solution of hydrochloric acid yields an aqueous solution of zinc chloride plus hydrogen gas<span>
</span>
Answer:
Explanation:
- For the balanced reaction:
<em>4Fe(s) + 3O₂(g) → 2Fe₂O₃(s).</em>
It is clear that 4 mol of Fe react with 3 mol of O₂ to produce 2 mol of Fe₂O₃.
- Firstly, we need to calculate the no. of moles of 35.8 grams of Fe metal:
no. of moles of Fe = mass/molar mass = (35.8 g)/(55.845 g/mol) = 0.64 mol.
- Now, we can find the no. of moles of O₂ is needed to react with the proposed amount of Fe:
<em><u>Using cross multiplication:</u></em>
4 mol of Fe is needed to react with → 3 mol of O₂, from stichiometry.
0.64 mol of Fe is needed to react with → ??? mol of O₂.
∴ The no. of moles of O₂ needed = (3 mol)(0.64 mol)/(4 mol) = 0.48 mol.
- Finally, we can get the volume of oxygen using the information:
<em>It is known that 1 mole of any gas occupies 22.4 L at standard P and T (STP).</em>
<em></em>
<em><u>Using cross multiplication:</u></em>
1 mol of O₂ occupies → 22.4 L, at STP conditions.
0.48 mol of O₂ occupies → ??? L.
∴ The no. of liters of O₂ = (0.48 mol)(22.4 L)/(1 mol) = 10.752 L.
