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3 years ago

The following reaction shows sodium hydroxide reacting with sulfuric acid.

Chemistry

2 answers:

amm18123 years ago

6 0

its answer is 35.5 grams

OverLord2011 [107]3 years ago

4 0

Answer: The mass of $Na_2SO_4$ produced will be 17.8 grams.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

For Sodium hydroxide:

Given mass of sodium hydroxide = 10 g

Molar mass of $NaOH=[(1\times 22.989)+(1\times 15.999)+(1\times 1.008)]g/mol=39.996g/mol$

Putting values in above equation, we get:

$\text{Moles of sodium hydroxide}=\frac{10g}{39.996g/mol}=0.25mol$

For the given chemical equation:

$2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O$

By Stoichiometry of the reaction:

2 moles of sodium hydroxide produces 1 mole of sodium sulfate.

So, 0.25 moles of sodium hydroxide will produce = $\frac{1}{2}\times 0.25=0.125moles$ of sodium sulfate.

Now, to calculate the mass of sodium sulfate, we use equation 1:

Moles of sodium sulfate = 0.125 moles

Molar mass of $Na_2SO_4=[(2\times 22.989)+(1\times 32.065)+(4\times 15.999)]g/mol=142.039g/mol$

Putting values in equation 1, we get:

$0.125mol=\frac{\text{Mass of }Na_2SO_4}{142.039g/mol}\\\\\text{Moles of }Na_2SO_4=17.8g$

Hence, the mass of $Na_2SO_4$ produced will be 17.8 grams.

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The fact that HBO2, a reactive compound, was produced rather than the relatively inert B2O3 was a factor in the discontinuation

Reil [10]

Answer:

1027.62 g

Explanation:

For $B_2H_6$ :-

Mass of $B_2H_6$ = 296.1 g

Molar mass of $B_2H_6$ = 27.66 g/mol

The formula for the calculation of moles is shown below:

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

Thus,

$Moles= \frac{296.1\ g}{27.66\ g/mol}$

$Moles\ of\ B_2H_6= 10.705\ mol$

From the balanced reaction:-

$B_2H_6(g) + 3 O2_{(l)}\rightarrow 2 HBO_2_{(g)}+ 2 H_2O_{(l)}$

1 mole of $B_2H_6$ react with 3 moles of oxygen

Thus,

10.705 mole of $B_2H_6$ react with 3*10.705 moles of oxygen

Moles of oxygen = 32.115 moles

Molar mass of oxygen gas = 31.998 g/mol

Mass = Moles * Molar mass = 32.115 * 31.998 g = 1027.62 g

8 0

3 years ago

Which group contains elements that have the following characteristics:

alekssr [168]

Answer:

Explanation:

the answer to you question Is 1)

6 0

3 years ago

Read 2 more answers

A sample of methane gas, CH4, occupies 3.25 L at temperature of 19.0 o C. If the pressure is held constant, what will be the tem

Artemon [7]

Answer:

625.46 °C

Explanation:

We'll begin by converting 19 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

T(°C) = 19 °C

T(K) = 19 °C + 273

T(K) = 292 K

Next, we shall determine the Final temperature. This can be obtained as follow:

Initial volume (V₁) = 3.25 L

Initial temperature (T₁) = 292 K

Final volume (V₂) = 10 L

Final temperature (T₂) =?

V₁/T₁ = V₂/T₂

3.25 / 292 = 10 / T₂

Cross multiply

3.25 × T₂ = 292 × 10

3.25 × T₂ = 2920

Divide both side by 3.25

T₂ = 2920 / 3.25

T₂ = 898.46 K

Finally, we shall convert 898.46 K to celsius temperature. This can be obtained as follow:

T(°C) = T(K) – 273

T(K) = 898.46 K

T(°C) = 898.46 – 273

T(°C) = 625.46 °C

Therefore the final temperature of the gas is 625.46 °C

4 0

3 years ago

Two chemist are trying to determine if their water sample is a mixture or pure substance. In their investigation they first filt

Effectus [21]

The correct answer is option B. Dirty water is a mixture of solid particles and liquid. It is both a mixture and pure substance.

The dirty water sample has both gravel and liquid water in it. After filtration the gravel is removed so the water sample looks clearer than before filtration. Liquid water is a pure substance because it is a compound that is made up of elements hydrogen and oxygen. Now the gravel is only physically combined with the liquid water, thus giving the water sample properties of a mixture. And like any mixture, gravel is physically separated through filtration from the liquid water.

Thus the water sample of the chemists is both a mixture and pure substance.

8 0

4 years ago

How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos

Alexeev081 [22]

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of $H_3PO_4$ = 54.219 g

Number of atoms of $Mg$ = $7.179\times 10^{23}$

Molar mass of $H_3PO_4$ = 98 g/mol

First we have to calculate the moles of $H_3PO_4$ and $Mg$ .

$\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}$

$\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol$

and,

$\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2$

From the balanced reaction we conclude that

As, 3 mole of $Mg$ react with 2 mole of $H_3PO_4$

So, 0.553 moles of $Mg$ react with $\frac{2}{3}\times 0.553=0.369$ moles of $H_3PO_4$

From this we conclude that, $H_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $Mg$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2$

From the reaction, we conclude that

As, 3 mole of $Mg$ react to give 3 mole of $H_2$

So, 0.553 mole of $Mg$ react to give 0.553 mole of $H_2$

Now we have to calculate the volume of $H_2$ gas at STP.

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas

So, 0.553 mole of hydrogen gas occupies $0.553\times 22.4=12.4L$ volume of hydrogen gas

Therefore, the volume of hydrogen gas produced will be, 12.4 L

4 0

3 years ago