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Ilya [14]
2 years ago
5

Suppose we have 100 g of each of the following substances. Which sample contain the greatest number of moles (F.W. = Formula Wei

ght) A. HCI, F.W. = 36.5 B. H_2O, F.W. = 18.0 C. MgCO_3 F.W. = 84.3 D. AlCI_3, F.W. = 133.3 E. NaCl, F.W. = 58.4
Chemistry
1 answer:
Sindrei [870]2 years ago
3 0

Answer:

100 g of water has the highest number of moles

Explanation:

Recall that the number of moles is obtained as given mass/formula weight

For HCl;

number of moles = 100g/36.5g/mol = 2.7 moles

For H2O;

number of moles = 100g/18g/mol = 5.5 moles

For MgCO3

number of moles = 100g/84.3 g/mol = 1.2 moles

For AlCl3

number of moles = 100g/133.3g/mol = 0.75 moles

For NaCl

number of moles = 100g/58.4 g/mol = 1.7 moles

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PLeaseee ITS Timedddd + BrainlIESTT
zmey [24]

Answer:

d. 12.3 grams of Al2O3

Explanation:

Based on the reaction:

4Al + 3O2 → 2Al2O3

<em>Where 4 moles of Al reacts in excess of oxygen to produce 2 moles of aluminium oxide.</em>

<em />

To solve this question we must find the moles of Aluminium. With these moles we can find the moles of aluminium oxide using the reaction:

<em>Moles Al -Molar mass: 26.9815g/mol-</em>

6.50g * (1mol / 26.9815g) = 0.241 moles Al

<em>Mass Al₂O₃ -Molar mass: 101.96g/mol-</em>

0.241 moles Al * (2 mol Al2O3 / 4 mol Al) = 0.120 moles Al2O3

0.120 moles Al2O3 * (101.96g / mol) =

12.3g of Al2O3 are produced.

Right answer is:

<h3>d. 12.3 grams of Al2O3 </h3>

8 0
3 years ago
What happens when carbon dioxide is passed through lime water for a long time? write with equation​
Vlada [557]

Answer:

when CO2 gas is passed through lime water it turns milky due to the formation of calcium carbonate which formula is CaCO3.

Ca(OH)2+ CO2------ CaCO3

when excess of carbon dioxide is passed through calcium carbonate calcium hydrogen carbonate is formed and solution become colourless.

CaCO3+CO2------ Ca(HCO3)

5 0
3 years ago
How many grams of CO2 are used when 6.0 g of O2 are produced? Express your answer with the appropriate units.
Ipatiy [6.2K]

Answer:

that is why co2 is in the power of 2ik

8 0
3 years ago
What is the mass (in grams) of 9.27x1024 molecules of methanol(CH3OH)
garri49 [273]
Moles of methanol = 9.27x10^24/6.02x10^23 = 15.398 moles.

Mass of methanol = moles of methanol x molar mass of methanol
                               = 15.398 x 32.042
                               = 493.38 grams.

Hope this helps!

5 0
3 years ago
Read 2 more answers
What expression approximates the volume of O2 consumed, measure at STP, when 55 g of Al reacts completely with excess O2?2 Al(s)
Genrish500 [490]

Answer:

34.28 L ( 1.5*22.4 L)

Explanation:

Calculation of the moles of aluminum as:-

Mass = 55 g

Molar mass of aluminum = 26.981539 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{55\ g}{26.981539\ g/mol}

Moles= 2.0384\ mol

According to the reaction:-

4Al+3O_2\rightarrow 2Al_2O_3

4 moles of aluminum react with 3 moles of oxygen gas

1 mole of aluminum react with \frac{3}{4} moles of oxygen gas

2.0384 moles of aluminum react with \frac{3}{4}\times 2.0384 moles of oxygen gas

Moles of oxygen gas = 1.5288 moles

At STP,  

Pressure = 1 atm  

Temperature = 273.15 K

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

1 atm × V = 1.5288 mol × 0.0821 L.atm/K.mol × 273.15 K

⇒V = 34.28 L ( 1.5*22.4 L)

7 0
3 years ago
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