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4 years ago

Consider the balanced equation for the following reaction:

Chemistry

1 answer:

mafiozo [28]4 years ago

3 0

Answer:

There will remain 78.9 grams of NH3.

Explanation:

Step 1: Data given

Mass CuO = 17.3 grams

Molar mass CuO = 79.545 g/mol

Mass NH3 = 81.4 grams

Molar mass NH3 = 17.03 g/mol

Step 2: The balanced equation

3CuO(s) + 2NH3(g) → 3H2O(l) + 3Cu(s) + N2(g)

Step 3: Calculate moles CuO

Moles CuO = mass CuO/ molar mass CuO

Moles CuO = 17.3 grams / 79.545 g/mol

Moles CuO = 0.217 moles

Step 4: Calculate moles NH3

Moles NH3 = 81.4 grams / 17.03 g/mol

Moles NH3 = 4.78 moles

Step 5: Calculate the limiting reactant

CuO is the limiting reactant. It will completely react. NH3 is in excess. There will react 2/3 * 0.217 = 0.145 moles. There will remain 4.78 - 0.145 = 4.635 moles NH3

Step 6: Calculate mass NH3 remaining

Mass NH3 = 4.635 moles * 17.03 g/mol

Mass NH3 = 78.9 grams

There will remain 78.9 grams of NH3.

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Explanation:

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3 years ago

A 200.0 mL solution of 0.40 M ammonium chloride was titrated with 0.80 M sodium hydroxide. What was the pH of the solution after

choli [55]

Answer:

9.25

Explanation:

Let first find the moles of $NH_4Cl$ and $NaOH$

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= 0.08 mole

number of moles of $NaOH$ = 0.80 mol/L × 50 × 10⁻³L

= 0.04 mole

The equation for the reaction is expressed as:

$NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}$

The ICE Table is shown below as follows:

$NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}$

Initial (M) 0.08 0.04 0

Change (M) - 0.04 -0.04 + 0.04

Equilibrium (M) 0.04 0 0.04

$K_a*K_b = 10^{-14} \ at \ 25^0C$

$K_a = \frac{10^{-14}}{K_b}$

$K_a = \frac{10^{-14}}{1.76*10^{-5}}$

$K_a= 5.68*10^{10}$

$pK_a = - log \ (K_a)$

$pK_a = - log \ (5.68*10^{-10})$

$pK_a = 9.25$

$pH = pKa + \ log (\frac{HB}{HA} )$ for buffer solutions

$pH = pKa + \ log (\frac{moles \ of \ base }{ moles\ of \ acid} )$ since they are in the same solution

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3 years ago

Consider the following generic chemical reaction:

jeka57 [31]

Answer:

The number of moles of B that will completely react with;

8 moles of A is 12 moles of B

9 moles of A is 13.5 moles of B

3 moles of A is 4.5 moles of B

10 moles of A is 15 moles of B

Explanation:

From the reaction, we have the mole ratio of the reactants is 2 moles of A combines with 3 moles of B to form 3 moles of C

Therefore, one mole of A which is 2 moles/2 will combine with 3/2 moles of B

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3 years ago