Answer:
39.6 g
Explanation:
The equation of the reaction is;
2Mg(s) + O2(g) --------> 2MgO(s)
To obtain the limiting reactant;
Number of moles in 26.4 g of Mg = 26.4g/24 g/mol = 1.1 moles
If 2 moles of Mg yields 2 moles of MgO
1.1 moles of Mg yields 1.1 * 2/2 = 1.1 moles of MgO
Number of moles in 26.4 g of O2 = 26.4 g/32g/mol = 0.825 moles
If 1 mole of O2 yields 2 moles of MgO
0.825 moles of O2 yields 0.825 moles * 2/1 = 1.65 moles of MgO
Hence Mg is the limiting reactant.
Theoretical yield of MgO = 1.1 moles of MgO * 40 g/mol = 44 g
Percent yield = 90%
Percent yield = actual yield/theoretical yield * 100
Actual yield = Percent yield * theoretical yield/100
Actual yield = 90 * 44/100
Actual yield = 39.6 g
the produce would be water and some form of salt
hope this helps
Answer:
(a) 1.92 moles of Bi produced.
(b) 80.6 grams
Explanation:
Balanced equation: Bi2O3(s) + 3C(s) → 2Bi(s) + 3CO(g)
1st find moles of Bi2O3:
Bi2O3 has Mr of 466 and mass of 447 g
2nd find moles of Bi:
Bi2O3 : 2Bi
→ 1 : 2 ------ this is molar ratio.
→ 0.959227 : (0.959227)*2
→ 0.959227 : 1.91845
→ 0.959227 : 1.92
Therefore 1.92 moles of Bi was produced.
3rd Find moles of 3CO:
Bi2O3 : 3CO
1 : 3
0.959227 : (0.959227 )*3
0.959227 : 2.87768
3CO has 2.87768 moles and we know the Mr is 28.
g
Therefore 80.575 grams of CO was produced.