If the actual yield of sodium chloride from the reaction of 8.3 g of sodium and 4.5 g of chlorine is 6.4 g, what is the percent yield?
Answer : 2Na(s) + Cl2(g) → 2NaCl(s)
Explanation: Friend me!!!
CaCO3(s) ⟶ CaO(s)+CO2(s)
<span>
moles CaCO3: 1.31 g/100 g/mole CaCO3= 0.0131 </span>
<span>
From stoichiometry, 1 mole of CO2 is formed per 1 mole CaCO3,
therefore 0.0131 moles CO2 should also be formed.
0.0131 moles CO2 x 44 g/mole CO2 = 0.576 g CO2 </span>
Therefore:<span>
<span>% Yield: 0.53/.576 x100= 92 percent yield</span></span>
I think the answer would be A because O is oxygen and it has 7. Although it’s in parentheses and has a 2 on the outside of those parentheses, so you would multiply and 7 x 2 = 14. 14 is larger than the other ones.
Hopefully I’m right and hopefully that helps.
True some explanations are not always based on empirical evidence
