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1 year ago

2 NH3 + 3 CuO →3 Cu + N₂ + 3 H₂O

Chemistry

1 answer:

Tatiana [17]1 year ago

5 0

40.1g of nitrogen gas is produced.

The equation given is

2 NH₃ + 3 CuO →3 Cu + N₂ + 3 H₂O

This equation is already balanced.

When 3 moles of CuO are consumed, 1 mole of nitrogen gas is produced.

We get 1 mole of nitrogen from 3 moles of copper oxide.

We need to find the number of moles of nitrogen gas produced when 4.3 moles of copper oxide are consumed.

4.3/3 x 1 = 1.433 mols

1.433 mols of nitrogen gas are produced
The molar mass of nitrogen gas is 14+14 = 28g
The amount of nitrogen gas produced in grams is 28x1.433 = 40.1g

40.1g of nitrogen gas can be made when 4.3 moles of CuO are consumed.

Learn more about molarity here:

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For 100.0 mL of a solution that is 0.040M CH3COOH and 0.010 M CH3COO, what would be the pH after adding 10.0 mL 50.0 mM HCl?

damaskus [11]

Answer:

The pH of the buffer is 3.90

Explanation:

The mixture of a weak acid CH3COOH and its conjugate base CH3COO produce a buffer that follows the equation:

pH = pKa + log [A-] / [HA]

Where pH is the pH of the buffer, pKa is the pKa of acetic acid (4.75), and [A-] could be taken as the moles of the conjugate base and [HA] the moles of thw weak acid.

To solve this question we need to find the moles of the CH3COOH and CH3COO- after the reaction with HCl:

CH3COO- + HCl → CH3COOH + Cl-

The moles of CH3COO- are its initial moles - the moles of HCl added

And moles of CH3COOH are its initial moles + moles HCl added

Moles CH3COO-:

Initial moles = 0.100L * (0.010mol / L) = 0.00100moles

Moles HCl = 0.010L * (0.050mol / L) = 0.000500 moles

Moles CH3COO- = 0.000500 moles

Moles CH3COOH:

Initial moles = 0.100L * (0.040mol / L) = 0.00400moles

Moles HCl = 0.010L * (0.050mol / L) = 0.000500 moles

Moles CH3COO- = 0.003500 moles

pH is:

pH = 4.75 + log [0.000500] / [0.00350]

pH = 3.90

<h3>The pH of the buffer is 3.90</h3>

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2 years ago