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beks73 [17]
3 years ago
5

Given the problem below, what would the first step be in solving it?

Chemistry
1 answer:
Hoochie [10]3 years ago
4 0

Answer:

First step would be convert to moles

Final Answer: 37.8 g of NaCl

Explanation:

The reaction is:

2Na + Cl₂ → 2NaCI

We convert the mass of each reactant to moles:

18 g . 1mol /23g = 0.783 moles of Na

23g . 1mol / 70.9g = 0.324 moles of chlorine

We use the mole ratio to determine the limiting reactant:

Ratio is 2:1. 2 moles of Na react to 1 mol of chlorine

Then, 0.783 moles of Na, may react to (0.783 . 1)/2 = 0.391 moles.

Excellent!. We need 0.391 moles of Cl₂ and we only have 0.324 moles available. That's why the Cl₂ is our limiting reactant.

We use the mole ratio again, with the product side. (1:2)

1 mol of Cl₂ can produce 2 moles of NaCl

Then, our 0.324 moles of gas, may produce (0.324 . 2)/1 = 0.648 moles

Finally, we convert the moles to grams:

0.648 mol . 58.45g/mol =

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The complete combustion of acetic acid, HC2H3O2(I) to form water, H2O(I), and CO2(g), at constant pressure releases 871.7 kJ of
lbvjy [14]

Answer:

HC2H3O2(I) + O2(g) ---> H2O(I) + CO2(g) ΔH= -72.35 kJ

Explanation:

We know that 5.0 g of acetic acid will contain, 5.0g/60 g/mol = 0.083 moles of acetic acid

Now from the reaction equation;

1 mole of acetic acid evolved -871.7 KJ of heat

0.083 moles of acetic acid will evolve 0.083 * -871.7 = -72.35 KJ

For 5.0 g of acetic acid, we can write;

HC2H3O2(I) + O2(g) ---> H2O(I) + CO2(g) ΔH= -72.35 kJ

3 0
3 years ago
How many moles of H2O are in 64.0 g of H2O
Alina [70]

Answer:

Explanation:

stoichiometry of C₂H₂ to H₂O is 2:2.

Number of moles of C₂H₂ = molar mass of C₂H₂  

Since the molar mass of C₂H₂  is 26 g/mol.

Number of C₂H₂  moles reacted = 64.0 g / 26 g/mol = 2.46 mol.

according to a molar ratio of 2:2.

the number of H₂O moles formed = a number of C₂H₂  moles reacted.

Therefore the number of H₂O moles produced = 2.46 mol

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Answer:

B

Explanation:

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