Answer:
A I did the exam and I saw A sorry if it is wrong dont have the best of memory
66.2% yield
The balanced equation for the reaction is:
2C6H10 + 17O2 ==> 12CO2 + 10H2O
So for every 2 moles of C6H10 consumed, we should get 12 moles of CO2. Or to simplify, for each mole of C6H10, we should get 6 moles of CO2. Now let's calculate the molar mass of C6H10 and CO2 and then determine how many moles of each we really have.
Atomic weight carbon = 12.0107
Atomic weight hydrogen = 1.00794
Atomic weight oxygen = 15.999
Molar mass C6H10 = 6*12.0107 + 10*1.00794 = 82.1436 g/mol
Molar mass CO2 = 12.0107 + 2*15.999 = 44.0087 g/mol
Moles C6H10 = 8.88 g / 82.1436 g/mol = 0.10810337 mol
Moles CO2 = 18.9 g / 44.0087 g/mol = 0.429460538 mol
Since we had 0.10810337 moles of C6H10, we should have gotten
6*0.10810337 = 0.648620221 moles of CO2, but only got 0.429460538 moles. So let's divide the actual yield by the theoretical yield to get the percentage yield.
0.429460538 / 0.648620221 = 0.662114014 = 66.2%
D since Francium has an atomic number (mass) of 87
Answer:
a) 2Mg + TiCl₄ ⟶ 2MgCl₂ + Ti
b) 1.02 kg
Step-by-step explanation:
a) Balanced equation
We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 24.30 47.87
2Mg + TiCl₄ ⟶ 2MgCl₂ + Ti
m/kg: 1.00
b) Mass of Mg
(i) Calculate the <em>moles of Ti
</em>
n = 1.00 kg Ti × (1 kmol Ti /47.87 kg Ti)
= 0.020 89 kmol Ti
(ii) Calculate the <em>moles of Mg
</em>
The molar ratio is (2 kmol Mg/1 kmol Ti)
n = 0.020 89 kmol Ti × (2 kmol Mg/1 kmol Ti)
= 0.041 78 kmol Mg
(iii) Calculate the <em>mass of Mg
</em>
m = 0.041 78 kmol Mg × (24.30 kg Mg/1 kmol Mg)
= 1.02 kg Mg
You need 1.02 kg Mg to produce 1.00 kg Ti.