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Anika [276]
3 years ago
6

Salicylic acid (C7H6O3) reacts with acetic anhydride (C4H6O3) to form acetylsalicylic acid (C9H8O4). 2C7H6O3(aq) + C4H6O3(aq) mc

002-1.jpg 2C9H8O4(aq) + H2O(l) What is the limiting reactant if 70.0 g of C7H6O3 and 80.0 g of C4H6O3 react?
Chemistry
2 answers:
olga nikolaevna [1]3 years ago
3 0
In order to understand what is the limiting reactant is useful to convert the masses of reactants into moles. The molecular masses are:

C_7H_6O_3 = 138\frac{g}{mol}\\ C_4H_6O_3 = 100\frac{g}{mol}

Converting masses into moles:

70\g \C_7H_6O_3\cdot \frac{1\ mol}{138\ g} = 0.5\ mol\ C_7H_6O_3

80\ g\ C_4H_6O_3\cdot \frac{1\ mol}{100\ g} = 0.8\ mol\ C_4H_6O_3

In the reaction, 2 moles of C_7H_6O_3 are necessary to react with 1 mol of C_4H_6O_3. Following this ratio:

0.8\ mol\ C_4H_6O_3\cdot \frac{2\ mol\ C_7H_6O_3}{1\ mol\ C_4H_6O_3} = 1.6\ mol\ C_7H_6O_3

We have only 0.5 mol of C_7H_6O_3, i.e, we don't have enough amount of salicylic acid, therefore, this is the limiting reactant.
ololo11 [35]3 years ago
3 0

Answer: B) Salicylic acid

Explanation: i just took the test !!!!!

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If 852.04 g Al(NO3)3 reacted with the 741.93 g Na2CO3, which would be the limiting reagent?
Sedaia [141]

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3. How many moles are 7.32 x 102<br> atoms of phosphorous?
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A buffer solution contains 0.496 M KHCO3 and 0.340 M K2CO3. If 0.0585 moles of potassium hydroxide are added to 250. mL of this
aev [14]

Answer:

pH = 10.9

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to say that the undergoing reaction between this buffer and OH⁻ promotes the formation of more CO₃²⁻ because it acts as the base, we can do the following:

n_{CO_3^{2-}}=0.34mol/L*0.250L+0.0585mol=0.1435mol\\\\n_{HCO_3^{-}}=0.34mol/L*0.250L-0.0585mol=0.0265mol

The resulting concentrations are:

[CO_3^{2-}]=\frac{0.1435mol}{0.25L}=0.574M \\

[HCO_3^{-}]=\frac{0.0265mol}{0.25L}=0.106M

Thus, since the pKa of this buffer system is 10.2, the change in the pH would be:

pH=10.2+log(\frac{0.574M}{0.106M} )\\\\pH=10.9

Which makes sense since basic OH⁻ ions were added.

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6 0
3 years ago
Copper oxide, CuO, reacts with hydrochloric acid, HCI, to produce copper chloride, CuCL2 and water
spayn [35]

Explanation:

El óxido de cobre (II), también llamado antiguamente óxido cúprico ({\displaystyle {\ce {CuO}}}{\displaystyle {\ce {CuO}}}), es el óxido de cobre con mayor número de oxidación. Como mineral se conoce como tenorita.

{\displaystyle {\ce {2Cu + O2 = 2CuO}}}{\displaystyle {\ce {2Cu + O2 = 2CuO}}}

Aquí, se forma junto con algo de óxido de cobre (I) como un producto lateral, por lo que es mejor prepararlo por calentamiento de nitrato de cobre (II), hidróxido de cobre (II) o carbonato de cobre (II):

{\displaystyle {\ce {2 Cu(NO3)2 = 2 CuO + 4 NO2+ O2}}}{\displaystyle {\ce {2 Cu(NO3)2 = 2 CuO + 4 NO2+ O2}}}

{\displaystyle {\ce {Cu(OH)2 (s) = CuO (s) + H2O (l)}}}{\displaystyle {\ce {Cu(OH)2 (s) = CuO (s) + H2O (l)}}}

{\displaystyle {\ce {CuCO3 = CuO + CO2}}}{\displaystyle {\ce {CuCO3 = CuO + CO2}}}

El óxido de cobre (II) es un óxido básico, así se disuelve en ácidos minerales tales como el ácido clorhídrico, el ácido sulfúrico o el ácido nítrico para dar las correspondientes sales de cobre (II):

{\displaystyle {\ce {CuO + 2 HNO3 = Cu(NO3)2 + H2O}}}{\displaystyle {\ce {CuO + 2 HNO3 = Cu(NO3)2 + H2O}}}

{\displaystyle {\ce {CuO + 2 HCl =CuCl2 + H2O}}}{\displaystyle {\ce {CuO + 2 HCl =CuCl2 + H2O}}}

{\displaystyle {\ce {CuO + H2SO4 = CuSO4 + H2O}}}{\displaystyle {\ce {CuO + H2SO4 = CuSO4 + H2O}}}

Reacciona con álcali concentrado para formar las correspondientes sales cuprato.

{\displaystyle {\ce {3 XOH + CuO + H2O = X3[Cu(OH)6]}}}{\displaystyle {\ce {3 XOH + CuO + H2O = X3[Cu(OH)6]}}}

Puede reducirse a cobre metálico usando hidrógeno o monóxido de carbono:

{\displaystyle {\ce {CuO + H2 = Cu + H2O}}}{\displaystyle {\ce {CuO + H2 = Cu + H2O}}}

{\displaystyle {\ce {CuO + CO = Cu + CO2}}}{\displaystyle {\ce {CuO + CO = Cu + CO2}}}

6 0
3 years ago
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