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3 years ago

How many moles of water are produced when 10 moles of hydrogen gas reacts with an excess of oxygen? (whole number)

Chemistry

1 answer:

asambeis [7]3 years ago

6 0

Answer:

12 moles H

Explanation:

Your tools of choice for stoichiometry problems will always be the mole ratios that exist between the chemical species that take part in the reaction.

As you know, the stoichiometric coefficients attributed to each compound in the balanced chemical equation can be thought of as moles of reactants needed or moles of products formed in the reaction.

In your case, the balanced chemical equation for this synthesis reaction looks like this

2(g]

→

(l]]

Notice that the reaction requires

moles of hydrogen gas and

mole of oxygen gas to produce

moles of water.

This tells you that the reaction produces twice as many moles of water as you have moles of oxygen gas that take part in the reaction.

You know that your reaction uses

6.0

moles of oxygen. Assuming that hydrogen gas is not a limiting reagent, you can say that the reaction will produce

6.0

moles O

⋅

moles H

moles O

12 moles H

Explanation:

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emmasim [6.3K]

Answer:

SUB TO HAVIN RX

Explanation:

8 0

3 years ago

Based upon the mass of baking soda (NaHCO3) and using an excess of HCl in this experiment, you will 1) determine the mass of CO2

weqwewe [10]

Equation of reaction

NaHCO3 + HCl ---------> NaCl + H2O + CO2(g)

1)The mass(actual yield) of CO2 can be gotten by isolating it from other products and getting its mass.

Assume 1.0g of CO2 was gotten as a product of the experiment

2) For the theoretical yield

The mass of NaHCO3 was not stated, but for the purpose of this solution, I would assume 2g of NaHCO3 was used for the experiment. You can substitute any parameter by following the steps I follow

Number of mole of NaHCO3 = Mass of NaHCO3/ Molar Mass of NaHCO3

Number of mole of NaHCO3 = 2/84.01 = 0.0238 moles

1 mole of NaHCO3 yielded 1 mole of CO2

0.0238 moles of NaHCO3 will yield 0.0238 moles of CO2

Mass of CO2 = Number of moles * Molar Mass

Mass of CO2 = 0.0238 * 44 = 1.0472g

Theoretical yield of CO2 = 1.0472 grams

3) Percentage yield of CO2 = Actual yield/Theoretical yield *100%

Percentage yield of CO2 = 1.0/1.0472 *100

Percentage yield of CO2 = 95.49%

8 0

4 years ago

Read 2 more answers

Definition of Solid State of Matter

Oksanka [162]

Solid state of matter
- the atoms are tightly packed in orderly form, they slightly vibrate but are stuck in a fixed position. Solids cannot be compressed and they maintain their shape

5 0

3 years ago

What is the volume at STP of 3.44 x 1023 molecules of CO2

almond37 [142]

Answer:

C. 12.8 liters.

Explanation:

The Standard Temperature and Pressure (STP) of a gas are 273.15 K and 100 kilopascals. From Avogadro's Law, a mole of carbon dioxide contains $6.022 \times 10^{23}$ molecules. If we suppose that carbon dioxide behaves ideally, then the equation of state for ideal gas is:

$P\cdot V = n\cdot R_{u}\cdot T$ (1)

$P\cdot V = \frac{r\cdot R_{u}\cdot T}{N_{A}}$ (1b)

Where:

$P$ - Pressure, measured in pascals.

$V$ - Volume, measured in liters.

$r$ - Amount of molecules, no unit.

$N_{A}$ - Avogadro's number, no unit.

$R_{u}$ - Ideal gas constant, measured in pascal-liters per mole-Kelvin.

$T$ - Temperature, measured in Kelvin.

If we know that $P = 100000\,Pa$ , $r = 3.44\times 10^{23}$ , $N_{A} = 6.022\times 10^{23}$ , $T = 273.15\,K$ and $R_{u} = 8.314\times 10^{3}\,\frac{L\cdot Pa}{mol\cdot K}$ , then the volume of carbon dioxide at STP is:

$V = \frac{r\cdot R_{u}\cdot T}{N_{A}\cdot P}$

$V = \frac{(3.44\times 10^{23})\cdot \left(8.314\times 10^{3}\,\frac{L\cdot Pa}{mol\cdot K} \right)\cdot (273.15\,K)}{(6.022\times 10^{23})\cdot (100000\,Pa)}$

$V = 12.972\,L$

Therefore, the correct answer is C.

8 0

3 years ago

■ How many moles of water are produced by burning 2.72 mol H2 in an excess of O2?

Karolina [17]

12 moles of h20 water

6 0

3 years ago