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3 years ago

If 24.1 g of sodium hydroxide react with 22.0 g of hydrochloric acid to form 35.3g

Chemistry

1 answer:

disa [49]3 years ago

4 0

Answer:

10.85 g of H2O.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

NaOH + HCl —> NaCl + H2O

Next, we shall determine the mass of NaOH that reacted and the mass of H2O produced from the balanced equation.

These can be obtained as illustrated below:

Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol

Mass of NaOH from the balanced equation = 1 × 40 = 40 g

Molar mass of H2O = (2×1) + 16 = 2 + 16 = 18 g/mol

Mass of H2O from the balanced equation = 1 × 18 = 18 g

Summary:

From the balanced equation above,

40 g of NaOH reacted to produce 18 g of H2O.

Finally, we shall determine the mass of water, H2O produced from the reaction as follow:

From the balanced equation above,

40 g of NaOH reacted to produce 18 g of H2O.

Therefore, 24.1 g of NaOH will react to produce = (24.1 × 18)/40 = 10.85 g of H2O.

Therefore, 10.85 g of H2O were produced from the reaction.

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The total volume of seawater is 1.5 x 1021L .Assume that seawater contains 3.1 percent sodium chloride by mass and that its dens

son4ous [18]

Answer:

Mass in kg = 4.7*10^19 kg

Mass in tons = 5.2*10^16 tons

Explanation:

<u>Given:</u>

Total volume of sea water = 1.5*10^21 L

Mass % NaCl in seawater = 3.1%

Density of seawater = 1.03 g/ml

<u>To determine:</u>

Total mass of NaCl in kg and in tons

<u>Calculation:</u>

Unit conversion:

1 L = 1000 ml

The volume of seawater in ml is:

$=\frac{1.5*10^{21}L*1000ml }{1L} =1.5*10^{24} ml$

$Mass\ seawater = Density*volume = 1.03g/ml*1.5*10^{24} ml=1.5*10^{24}g$

$Mass\ NaCl\ in\ seawater = \frac{3.1}{100}*1.5*10^{24} g=4.7*10^{22} g$

To convert mass from g to Kg:

1000 g = 1 kg

$Mass\ seawater(kg) = \frac{4.7*10^{22}g*1kg }{1000g} =4.7*10^{19} kg$

To convert mass from g to tons:

1 ton = 9.072*10^6 g

$Mass\ seawater(tons) = \frac{4.7*10^{22}g*1ton }{9.072*10^{6}g } =5.2*10^{16} tons$

5 0

3 years ago

How many moles of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas?

tino4ka555 [31]

Taking into account the reaction stoichiometry, 340.0 moles of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

CO₂ + 4 H₄ → CH₄ + 2 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

CO₂: 1 mole
H₄: 4 moles
CH₄: 1 mole
H₂O: 2 moles

<h3>Moles of CH₄ formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 1 mole of CO₂ form 4 moles of CH₄, 85.1 moles of CO₂ form how many moles of CH₄?

$moles of CH_{4} =\frac{85.1 moles of CO_{2}x4 moles of CH_{4} }{1 moles of CO_{2}}$

<u><em>moles of CH₄= 340.4 moles</em></u>

Then, 340.0 moles of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas

Learn more about the reaction stoichiometry:

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6 0

1 year ago

How many moles of carbon dioxide gas can be produced when 7.94 moles of benzene (C6H6) react with excess oxygen? Be sure to show

Sliva [168]

The balanced equation is

<span>2 C6H6 +15 O2 = 12 CO2 + 6 H2O </span>

<span>the ratio between C6H6 and CO2 is 2 : 12 </span>

<span>moles CO2 produced = 7.94 x 12 / 2 =47.6</span>

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Answer:

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