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.
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Answer is: volume will be 6,7 L.
Boyle's Law: the pressure volume law - <span> volume of a given amount of gas held varies inversely with the applied pressure when the temperature and mass are constant.
p</span>₁V₁ = p₂V₂.
90 kPa · 5 L = 67 kPa · V₂.
V₂ = 90 kPa · 5 L / 67 kPa.
V₂ = 6,7 L, but same amount of oxygen.
Wym kingdom? I dont get it
The mass of CO2 produced by 26g of acetylene is 88g.
Given ,
In an oxyacetylene torch, acetylene gas (ethyne; HCCH) burns to produce carbon dioxide and water vapour.
The acetylene combustion reaction is given by,
H2O + HCCH + 5/2 O=O 2CO2
Heat of reaction for acetylene combustion = 1259kj/mol
CO2 has a molecular mass of 44g/mol.
2 moles of CO2 have a molecular mass of 88g.
On combustion, 1 mole of acetylene yields 2 moles of CO2.
Thus, 26g of acetylene produces 88g of CO2 when burned.
As a result, the mass of carbon dioxide produced by 26g of acetylene is 88g.
Learn more about acetylene here :
brainly.com/question/15346128
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