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How many grams of H2O will be formed when 32.0 g H2 is mixed with 84.0 g of O2 and allowed to react to form water

adell [148]

Answer:

94.57 g H2O

Explanation:

32.0 g H2

Molar Mass H2: 2.01 g/mol

84.0 g O2

Molar Mass O2: 32.00 g/mol

Molar Mass H2O: 18.02 g/mol

First balance the eqution:

2H2 + O2 = 2H2O

Find the amount of moles that both 32.0 g of H2 and 84.0 g of O2 can produce using the molar masses:

32.0 g H2 x ((1 mole H2)/ (2.015 g H2)) = 15.88 moles H2

84.0 g O2 x ((1 mole O2)/(32.00 g O2)) = 2.63 moles O2

Now you can find the amount of grams of H2O each reactant will produce:

**keep in mind of mole ratios from balanced equation**

15.88 moles H2 x ((2 moles H2O)/(2 moles H2)) x ((18.015 g H2O)/(1 mole H2O) =286.07 g H2O

2.63 moles O2 x ((2 moles H2O)/(1 mole O2)) x ((18.015 g H2O)/(1 mole H2O) = 94.76 g H2O

94.76 is the final answer because this is the limiting reactant, meaning it produces less product than the H2 so it limits the reaction from producing anymore product from the amount calculated.

5 0

2 years ago

How many grams of ice at -12.1 ∘C can be completely converted to liquid at 15.2 ∘C if the available heat for this process is 4.4

ra1l [238]

Answer:

The mass of ice is 1,06 kg

Explanation:

Formula for calorimetry: Q = m . C . ΔT

Formula for phase transition, with latent heat: Q = L . m

m . C . ΔT + L . m + m . C . ΔT = 4490 kJ

m . 2,01 J/g°C (0° - (-12,1°C) + 334 J/g . m + m . 4,186 J/g°C . (15,2°C - 0°C)

Our unknown value is mass

You should look for latent heat of Ice and of course, specific heat of water.

Those are constant and they are known.

As the units are in J, and the heat for the process is in Kj we have to convert the final number into J

4490kJ . 1000 = 449000 J

m. 2,01 J/g°C (0°-(-12,1°C) + 334 J/g .m +m .4,186 J/g°C .(15,2°C - 0°C) = 449000 J

m. 24,321 J/g + 334 J/g .m + 63,62 J/g . m = 449000 J

421,94 J/g . m = 449000 J

m = 449000 J /421,94 g/J = 1064,1 g ---->1,06 kg

8 0

2 years ago

How does carbon dioxide behave differently (its properties) as a gas compared with as a solid?

MaRussiya [10]

is made up of a carbon atom covalently double bonded to two oxygen atoms. Carbon dioxide exists in Earth's atmosphere as a trace gas at a concentration of about 0.04 percent by volume.

5 0

2 years ago

Using the table below, what is the change in enthalpy for the following reaction? 3CO (g) + 2Fe2O3 (s) Imported Asset Fe(s) + 3C

zhuklara [117]

To solve this problem, we should recall that the change in enthalpy is calculated by subtracting the total enthalpy of the reactants from the total enthalpy of the products:

ΔH = Total H of products – Total H of reactants

You did not insert the table in this problem, therefore I will find other sources to find for the enthalpies of each compound.

ΔHf CO2 (g) = -393.5 kJ/mol

ΔHf CO (g) = -110.5 kJ/mol

ΔHf Fe2O3 (s) = -822.1 kJ/mol

ΔHf Fe(s) = 0.0 kJ/mol

Since the given enthalpies are still in kJ/mol, we have to multiply that with the number of moles in the formula. Therefore solving for ΔH:

ΔH = [3 mol ( − 393.5 kJ/mol) + 1 mol (0.0 kJ/mol)] − [3 mol ( − 110.5 kJ/mol) + 2 mol ( − 822.1 kJ/mol)]

ΔH = 795.2 kJ

3 0

2 years ago

How many moles of water are created if 3 moles of hydrogen react completely with excess oxygen?

Alenkinab [10]

Answer:

3.0 moles.

Explanation:

It is a stichiometry problem.

The chemical reaction of reacting hydrogen with oxygen to produce water is:

H₂ + 1/2 O₂ → H₂O.

It is clear that 1.0 mole of H₂ reacts with 0.5 mole of O₂ to produce 1.0 mole of water.

The ratio of the reacting hydrogen to the produced water is 1:1.

∴ The number of moles of water created from reacting 3.0 moles of hydrogen completely with excess oxygen = 3.0 moles.

4 0

2 years ago