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lord [1]
2 years ago
12

50 POINTS! PLEASE HELP!Solid sodium nitrate reacts with sulfuric acid to produce sodium sulfate and nitric acid. You have 36 gra

ms of sodium nitrate and 21 grams of sulfuric acid. What is the limiting reactant?

Chemistry
1 answer:
ch4aika [34]2 years ago
4 0
The limiting reagent (or limiting reactant or limiting agent) in a chemical reaction is a reactant that is totally consumed when the chemical reaction is completed. The amount of product formed is limited by this reagent, since the reaction cannot continue without it. If one or more other reagents are present in excess of the quantities required to react with the limiting reagent, they are described as excess reagents or excess reactants (xs).
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A solution was prepared by mixing 0.5000 m hno2 with 0.380 m no2-. ka = 4.58 x 10-4. calculate the ph of the solution. show work
adell [148]

pH of buffer can be calculated as:

pH=pKa+log[salt]/[Acid]

As ka = 4.58 x 10-4

Concentration of [Salt] that is NO2(-1)=0.380M

Concentration of [Acid] that is HNO2=0.500M

So, pH= -log(4.58*10^-4)+log((0.380)/0.500))

=3.21

So pH of solution will be 3.21

7 0
3 years ago
Chromium(III) oxide can be prepared by heating chromium(IV) oxide in vacuo at high temperature: 4Cr02 —2Cr2O3 +02 The reaction o
kkurt [141]

<u>Answer:</u> The theoretical yield and percent yield of chromium (III) oxide is 434.72 grams and 92.6 % respectively.

<u>Explanation:</u>

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

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

Given mass of CrO_2 = 480.1 g

Molar mass of CrO_2 = 84 g/mol

Putting values in equation 1, we get:

\text{Moles of }CrO_2=\frac{480.1g}{84g/mol}=5.72mol

For the given chemical equation:

4CrO_2\rightarrow 2Cr_2O_3+O_2

By Stoichiometry of the reaction:

4 moles of CrO_2 produces 2 moles of chromium (III) oxide

So, 5.72 moles of CrO_2 will produce = \frac{2}{4}\times 5.72=2.86mol of chromium (III) oxide

Now, calculating the mass of chromium (III) oxide from equation 1, we get:

Molar mass of chromium (III) oxide = 152 g/mol

Moles of chromium (III) oxide = 2.86 moles

Putting values in equation 1, we get:

2.86mol=\frac{\text{Mass of chromium (III) oxide}}{152g/mol}\\\\\text{Mass of chromium (III) oxide}=(2.86mol\times 152g/mol)=434.72g

To calculate the percentage yield of chromium (III) oxide, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of chromium (III) oxide = 402.4 g

Theoretical yield of chromium (III) oxide = 434.72 g

Putting values in above equation, we get:

\%\text{ yield of chromium (III) oxide}=\frac{402.4g}{434.72g}\times 100\\\\\% \text{yield of chromium (III) oxide}=\%

Hence, the theoretical yield and percent yield of chromium (III) oxide is 434.72 grams and 92.6 % respectively.

7 0
3 years ago
He generic metal a forms an insoluble salt ab(s) and a complex ac5(aq). the equilibrium concentrations in a solution of ac5 were
elena55 [62]
Balanced chemical reaction: A + 5C ⇄ AC₅.
<span>[A] = 0.100 M; equilibrium concentration.
</span><span>[C] = 0.0380 M.
</span>[AC₅] = 0.100 M.
Kf = [AC₅] / ([A] · [C]⁵).
Kf = 0.100 M ÷ (0.100 M · (0.0380 M)⁵.
Kf = 12620658.54 = 1,26·10⁷.
<span>The formation constant can be calculated when </span>chemical equilibrium is reached, when the forward reaction rate is equal to the reverse reaction rate.
7 0
3 years ago
How much sugar does a bottle of sparkling water have help me plz it for a science project
zubka84 [21]

Answer:

it has 0g of sugar (hopes this helps)

3 0
3 years ago
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A chemical reaction in which compounds break up into simpler constituents is a _______ reaction.
klio [65]
C. decomposition reaction
8 0
3 years ago
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