¿Por qué se unen los átomos?
2 answers:
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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.
Answer:
2 water + sugar + lemon juice → 4 lemonade
Moles of water present in 946.36 g of water=\frac{946.36 g}{236.59 g/mol}=4 mol=
236.59g/mol
946.36g
=4mol
Moles of sugar present in 196.86 g of water=\frac{196.86 g}{225 g/mol}=0.8749 mol=
225g/mol
196.86g
=0.8749mol
Moles of lemon juice present in 193.37 g of water=\frac{193.37 g}{257.83 g/mol}=0.7499 mol=
257.83g/mol
193.37g
=0.7499mol
Moles of lemonade in 2050.25 g of water=\frac{2050.25 g}{719.42 g/mol}=2.8498 mol=
719.42g/mol
2050.25g
=2.8498mol
As we can see that number of moles of lemon juice are limited.
So, we will consider the reaction will complete in accordance with moles of lemon juice.
1 mole lemon juice reacts with 2 mol of water,then 0.7499 mol of lemon juice will react with:
\frac{2}{1}\times 0.7499 mol = 1.4998 mol
1
2
×0.7499mol=1.4998mol of water
Mass of water used = 1.4998 mol × 236.59 g/mol=354.8376 g
Water remained unused = 946.36 g - 354.8376 g =591.5223 g
1 mole lemon juice reacts with mol of sugar,then 0.7499 mol of lemon juice will react with:
\frac{1}{1}\times 0.7499 mol = 0.7499 mol
1
1
×0.7499mol=0.7499mol of water
Mass of sugar used = 0.7499 mol × 225 g/mol = 168.7275 g
Sugar remained unused = 196.86 g - 28.1325 g
1 mole of lemon juice gives 4 moles of lemonade.
Then 0.7499 mol of lemon juice will give:
\frac{4}{1}\times 0.7499 mol=2.996 mol
1
4
×0.7499mol=2.996mol of lemonade
Mass of lemonade obtained = 2.996 mol × 719.42 g/mol = 2157.9722 g
Theoretical yield of lemonade = 2157.9722 g
Experimental yield of lemonade = 2050.25 g
Percentage yield of lemonade:
\frac{\text{Experimental yield}}{\text{theoretical yield}}\times 100
theoretical yield
Experimental yield
×100
\frac{2050.25 g}{2157.9722 g}\times 100=95.00\%
2157.9722g
2050.25g
×100=95.00%
