Answer:
Percent yield of SiC is 77.0%.
Explanation:
Balanced reaction: 
Molar mass of SiC = 40.11 g/mol
Molar mass of 
= 60.08 g/mol
So, 100.0 kg of = 
moles of = 1664 moles of
According to balanced equation, 1 mol of produces 1 mol of SiC
Therefore, 1664 moles of produce 1664 moles of SiC
Mass of 1664 moles of SiC = 
= 66743g = 66.74 kg (4 sig. fig.)
Percent yield of SiC = [(actual yield of SiC)/(theoretical yield of SiC)]
%
= 
%
= 77.0%
<span>Answer: 100 ml
</span>
<span>Explanation:
1) Convert 1.38 g of Fe₂S₃ into number of moles, n
</span>i) Formula: n = mass in grass / molar mass
<span>
ii) molar mass of </span><span>Fe₂S₃ =2 x 55.8 g/mol + 3 x 32.1 g/mol = 207.9 g/mol
</span>
iii) n = 1.38 g / 207.9 g/mol = 0.00664 moles of <span>Fe₂S₃
</span>
<span>2) Use the percent yield to calculate the theoretical amount:
</span>
<span>65% = 0.65 = actual yield/ theoretical yield =>
</span>theoretical yield = actual yield / 0.65 = 0.00664 moles / 0.65 = 0.010 mol <span>Fe₂S₃</span><span>
3) Chemical equation:
</span>
<span> 3 Na₂S(aq) + 2 FeCl₃(aq) → Fe₂S₃(s) + 6 NaCl(aq)
4) Stoichiometrical mole ratios:
</span>
<span>3 mol Na₂S : 2 mol FeCl₃ : 1 mol Fe₂S₃ : 6 mol NaCl
5) Proportionality:
</span>2moles FeCl₃ / 1 mol Fe₂S₃ = x / 0.010 mol Fe₂S₃
<span>
=> x = 0.020 mol FeCl₃
6) convert 0.020 mol to volume
</span>
<span>i) Molarity formula: M = n / V
</span>
<span>ii) V = n / M = 0.020 mol / 0.2 M = 0.1 liter = 100 ml
</span>
<span>Mol is the unit of amount of substance. It is equal to 6.02 x 10^23 molecules. Now, One mol of Sodium chloride (NaCl) contains 6.022x 10^23 molecules of NaCl. Also, the number atoms of both Na (sodium) and Cl (chlorine) will be equal. Similatly, One mol of Aluminium Chloride (AlCl3) contains 6.022x 10^23 molecules of (AlCl3) but the ratio of Al and Cl atoms will be 1:3</span>
O = C = O Straight because there is no solitary electrons on C
