If you were to use 3.50 mL of 0.250 M copper(II) sulfate in an experiment similar to the one you just completed, what volume of 0.250 M sodium hydroxide would you need so that neither was a limiting

reactant? Justify your answer.

A tube containing 8.00 mL of 0.250 M copper(II) sulfate (CuSOs) contains 2.00 x 10 mole of

coppen II) sulfate. How many grams of solid copper(II) sulfate would this be equivalent to? Show

your work using dimensional analysis.

6.

How many moles of sodium hydroxide would be needed for a complete reaction with 4.80 x 10- mole

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11

a. volume of NaOH needed = 0.07 L or 7.0 mL since mole ratio of CuSO₄ to NaOH is 1 : 2

b. mass of copper (ii) sulfate = 0.32 g

c. Number of moles of NaOH needed, Y = 9.6 * 10⁻¹¹ moles of NaOH

Explanation:

The equation of the reaction is as follows:

CuSO₄ (aq) + 2NaOH(aq) —> Na₂SO₄ (aq) + Cu(OH)₂(s)

Number of moles of CuSO₄ in 3.50 mL of 0.250 M solution

number of moles = molarity * volume

number of moles = 0.250 M * 3.50 mL *  1L/1000 mL = 0.000875 moles

since mole ratio of CuSO₄ to NaOH is 1 : 2,

number of moles of NaOH required = 2 * 0.000875 = 0.00175moles

volume of NaOH required = number of moles/molarity

volume = 0.00175/0.250

volume = 0.07 L or 7.0 mL

b. number of moles = molarity * volume

number of moles = mass/molar mass

molar mass of CuSO₄ = 160 g/mol

number of moles = 0.250 mol/L * 8.00 mL * 1 L/1000 mL

number of moles = 2.00 * 10⁻² moles

2.00 * 10⁻² moles = mass/160 g/mol

mass = 2.00 * 10⁻² moles * 160 g/mol

mass = 0.32 g

c. Mole ratio of CuSO₄ to NaOH is 1 : 2

let number of moles of NaOH be Y

1 mole of CuSO₄/2 moles of NaOH= 4.8 * 10⁻¹¹ moles of CuSO₄/ Y

Y = 2 moles of NaOH * 4.8 * 10⁻¹¹ moles of / 1 moles of CuSO₄

Y = 9.6 * 10⁻¹¹ moles of NaOH

the only liquid elements at standard temperature and pressure are bromine (br) and mercury (hg). although, elements caesium (cs), rubidium (rb), francium (fr) and gallium (ga) become liquid at or just above room temperature.

explanation: