Answer:
46.0g of Iron are produced
Explanation:
Based on the chemical reaction:
FeO(l) + Mg(l) → Fe(l) + MgO(s)
<em>1 mole of Iron (II) oxide reacts per mole of Mg to produce 1 mole of iron</em>
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To solve this question we need to convert each mass of reactant to moles using its respectives molar masses in order to find limitng reactant. Moles of limiting reactant = Moles of iron produced:
<em>Moles FeO (Molar mass: 71.85g/mol):</em>
80.0g * (1mol / 71.85g) = 1.11moles FeO
<em>Moles Mg (Molar mass: 24.305g/mol)</em>
20.0g * (1mol / 24.305g) = 0.823 moles Mg
As moles of Mg < Moles FeO, Mg is limiting reactant and the moles of Fe are 0.823 moles.
The mass of Iron produced is:
0.823 moles Fe * (55.845g/mol) =
46.0g of Iron are produced
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The amount of atoms in each level and the positive and negative atoms in each level also
Answer:
1326.6 × 10⁻²³g
Explanation:
Given data:
Number of atoms of iron = 143 atoms
Average atomic mass of iron = 55.845 amu
Mass of iron = ?
Solution:
It is known that,
55.845 g of iron = one mole of iron = 6.02× 10²³ atoms
For 143 atoms:
143 atoms /6.02× 10²³ atoms × 55.845 g
23.754 × 10⁻²³× 55.845 g
1326.6 × 10⁻²³g
Answer:
1.72 moles of H₂
Explanation:
The balanced equation
2Na+ 2H₂O → <u>2</u>NaOH + <u>1C</u>
tells us we'll get <u>1</u> mole of H₂ for every <u>2</u> moles of NaOH. We can express this as a molar ratio: (1 mole H₂)/(2 moles NaOH).
The reaction produces 138 g of NaOH. Divide this mass by the molar mass of NaOH (39.99 g/mole)to obtain moles NaOH:
138g NaOH/(40 g/mole) = 3.45 moles of NaOH.
To find the moles H₂ produced, we can multiply the moles NaOH by the molar ratio from above:
(3.45 moles NaOH)*((1 mole H₂)/(2 moles NaOH) = (3.45/2) moles H₂. [Note how the "moles NaOH" unit cancels, leaving just "moles H₂"]
We should produce 1.72 moles of H₂.
