Answer:
0.6258 g
Explanation:
To determine the number grams of aluminum in the above reaction;
- determine the number of moles of HCl
- determine the mole ratio,
- use the mole ratio to calculate the number of moles of aluminum.
- use RFM of Aluminum to determine the grams required.
<u>Moles </u><u>of </u><u>HCl</u>
35 mL of 2.0 M HCl
2 moles of HCl is contained in 1000 mL
x moles of HCl is contained in 35 mL
We have 0.07 moles of HCl.
<u>Mole </u><u>ratio</u>
6HCl(aq) + 2Al(s) --> 2AlCl3(aq) + 3H2(g)
Hence mole ratio = 6 : 2 (HCl : Al
- but moles of HCl is 0.07, therefore the moles of Al;
Therefore we have 0.0233333 moles of aluminum.
<u>Grams of </u><u>Aluminum</u>
We use the formula;
The RFM (Relative formula mass) of aluminum is 26.982g/mol.
Substitute values into the formula;
The number of grams of aluminum required to react with HCl is 0.6258 g.
The mass (in grams) of iron, Fe that can be made from 21.5 g of Fe₂O₃ is 15.04 g
We'll begin by writing the balanced equation for the reaction. This is given below:
2Fe₂O₃ -> 4Fe + 3O₂
- Molar mass of Fe₂O₃ = 159.7 g/mol
- Mass of Fe₂O₃ from the balanced equation = 2 × 159.7 = 319.4 g
- Molar mass of Fe = 55.85 g/mol
- Mass of Fe from the balanced equation = 4 × 55.85 = 223.4 g
From the balanced equation above,
319.4 g of Fe₂O₃ decomposed to produce 223.4 g of Fe
<h3>How to determine the mass of iron, Fe produced</h3>
From the balanced equation above,
319.4 g of Fe₂O₃ decomposed to produce 223.4 g of Fe
Therefore,
21.5 g of Fe₂O₃ will decompose to produce = (21.5 × 223.4) / 319.4 = 15.04 g of Fe
Thus, 15.04 g of Fe were produced.
Learn more about stoichiometry:
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Answer:
D. 4
Explanation:
Answer and Explanation: Carbon can form a maximum of four covalent bonds. Carbon can share up to four pairs of electrons, therefore, the carbon atom fills its outer energy level and achieves chemical stability.
<span>Well if you're looking for grams, all you need to do is cancel out units.
(ml)(g/ml)=g because the ml cancels out.
Thus, multiply: (85.32ml)(1.03g/ml)=...I'll let you solve this. :)
Good luck! Hope that helped. When in doubt, look at the units.</span>
