If we burn 1.5 g of iron, how many grams of iron (III) oxide will be produced?
The equation is 4Fe (s) + 3 O2 (g) ==> 2 Fe2O3
Answers
Answer:
2.145g Fe2O3
Explanation:
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Fe atomic mass:55.8u
O atomic mass :16u
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As per the balanced equation,
- 4Fe ==> 2Fe2O3
- 4×55.8g Fe==>2((2×55.8)+(3×16))g Fe2O3
- 223.2g Fe ==>319.2g Fe2O3
- 1g Fe. ==>319.2/223.2 g Fe2O3
- 1g Fe. ==>1.43g Fe2O3
- 1.5g Fe. ==> 1.43×1.5g Fe2O3
==> 2.145g Fe2O3
i,e. 1.5 g Fe produces 2.145 g Fe2O3
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hope it helps...!!thank you...!!
Question : If we burn 1.5g of iron, how many grams of iron (III) oxide will be produced ?
Solution : weight of iron = 1.5 g
atomic weight of iron = 56 g/mol
so, no of moles of iron = 1.5/56 = 0.027 mol
The equation is 4Fe + 3O₂ => 2Fe₂O₃
from equation, it is clear that 4 moles of iron produce 2 moles of iron(III) oxide.
so, 0.027 mole of iron will produce 2/4 × 0.027 = 0.0135 mol of iron (III) oxide.
mass of iron (III) oxide = mole × molecular weight of Fe₂O₃
= 0.0135 mol × (56 × 2 + 16 × 3)
= 0.0135 × (112 + 48)
= 0.0135 × 160
= 2.16 grams.
Therefore mass of iron(III) oxide is 2.16 grams.
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