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Harrizon [31]
3 years ago
6

How many grams of water can be formed when 80 grams of sodium hydroxide (NaOH) reacts with an excess of sulfuric acid, H2SO4, in

solution?
2NaOH + H2SO4 --> Na2SO4 + 2H2O

_____ grams water
Chemistry
1 answer:
podryga [215]3 years ago
5 0

Answer:

36g of H2O.

Explanation:

We'll begin by writing the balanced equation for the reaction:

2NaOH + H2SO4 —> Na2SO4 + 2H2O

Next, we shall determine the mass of NaOH that reacted and the mass of H2O produced from the balanced equation. This is illustrated below:

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

Mass of NaOH from the balanced equation = 2 x 40 = 80g

Molar mass of H2O = (2x1) + 16 = 18g/mol

Mass of H2O from the balanced equation = 2 x 18 = 36g.

From the balanced equation above, we can see evidently that:

80g of NaOH reacted to produce 36g of H2O.

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Given the following heats of combustion. CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ C(graphite) + O2(g) CO2(g) ΔH
sergeinik [125]

Answer:

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation:

The formation reaction of CH_3OH will be,

C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g),\Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

C(graphite)+O_2(g)\rightarrow CO_2(g), \Delta H_1=-393.5kJ/mole..[1]

H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l), \Delta H_2=-285.8kJ/mole..[2]

CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l) , \Delta H_3=-726.4kJ/mole..[3]

Now we will reverse the reaction 3, multiply reaction 2 by 2  then adding all the equations, Using Hess's law:

We get :

C(graphite)+O_2(g)\rightarrow CO_2(g) , \Delta H_1=-393.5kJ/mole..[1]

2H_2(g)+2O_2(g)\rightarrow 2H_2O(l) ,\Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol..[2]

CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g) ,\Delta H_3=726.4kJ/mole [3]

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+2\times \Delta H_2+\Delta H_3

\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)

\Delta H=-238.7kJ/mole

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

4 0
3 years ago
blood plasma for average adults has a volume of 3100mL. if the density of plasma is 1.020 g/mL, what mass of plasma is in the av
Gnoma [55]

Answer:

The volume of blood plasma in adults is 3.1 L. it's density is 1.03 g/cm3.

Explanation:

4 0
3 years ago
An iron nail rusts when exposed to oxygen. According to the following reaction, how many moles of iron(III) oxide will be formed
Zigmanuir [339]

Answer:

0.453 moles

Explanation:

The balanced equation for the reaction is:

2Fe(s) + 3O2(g)  ==>  2Fe2O3

From the equation,  mass of O2 involved = 16 x 2 x 3 = 96g

                                 mass of Fe2O3 involved = [(2x26) + 3 x 16] x 2

                                                                            = 100g

                Therefore 96g of O2 produced 100g of Fe2O3

                                  32.2g of O2 Will produce   100x32.2/96

                                                   = 33.54g of Fe2O3

Converting it to mole using   number of mole = mass/molar mass

but molar mass of Fe2O3 = 26 + (16 X 3)

                                           = 74g/mole

Therefore number of mole of 33.54g of Fe2O3 = 33.54/74

                                                                           = 0.453 moles

5 0
3 years ago
I NEED HELP ASAP!!!
Rus_ich [418]

Explanation:

To answer this question, we'll need to use the Ideal Gas Law:

p

V

=

n

R

T

,

where

p

is pressure,

V

is volume,

n

is the number of moles

R

is the gas constant, and

T

is temperature in Kelvin.

The question already gives us the values for

p

and

T

, because helium is at STP. This means that temperature is

273.15 K

and pressure is

1 atm

.

We also already know the gas constant. In our case, we'll use the value of

0.08206 L atm/K mol

since these units fit the units of our given values the best.

We can find the value for

n

by dividing the mass of helium gas by its molar mass:

n

=

number of moles

=

mass of sample

molar mass

=

6.00 g

4.00 g/mol

=

1.50 mol

Now, we can just plug all of these values in and solve for

V

:

p

V

=

n

R

T

V

=

n

R

T

p

=

1.50 mol

×

0.08206 L atm/K mol

×

273.15 K

1 atm

= 33.6 L

this is not the answer but it will help you

do by the formula it is on the answer

3 0
3 years ago
The column just before the stable noble gases are called the
Sergio039 [100]
Is called the Halogen
6 0
3 years ago
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