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grigory [225]
3 years ago
14

The student's lab manual says to mix some of his Na2CO3 solution with an aqueous solution of copper(II) sulfate (CuSO4)

Chemistry
1 answer:
lord [1]3 years ago
3 0

Explanation:

When the student mixed the solution sodium carbonate with solution of copper(II) sulfate ; Copper Hydroxocarbonate , sodium sulfate and carbon dioxide gas was obtained as a products.

The balanced chemical reaction

2Na_2CO_3+2CuSO_4\rightarrow Cu_2(OH)_2CO_3+2Na_2SO_4+CO_2

Where:

Cu_2(OH)_2CO_3 = Copper(II) Hydroxocarbonate

Na_2CO_3 = Sodium carbonate

CuSO_4 = Copper(II) sulfate

Na_2SO_4 = Sodium sulfate

CO_2 = Carbon-dioxide

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4 0
2 years ago
PLEASE HELP DUE IN EXACTLY 15 mins!! i will give you branliest
Deffense [45]

2032533 \sq22222rt[2]{?}

4 0
3 years ago
Help?
tino4ka555 [31]

Answer:

There is 52.33 grams of water produced.

Explanation:

Step 1: Data given

Mass of propane burned = 32.00 grams

Molar mass of propane = 44.1 g/mol

Oxygen is in excess

Molar mass of water = 18.02 g/mol

Step 2: The balanced equation

C3H8 + 5O2 → 4H2O + 3CO2

Step 3: Calculate moles of propane

Moles of propane = mass propane / molar mass of propane

Moles of propane = 32.00 grams / 44.1 g/mol

Moles of propane = 0.726 moles

Step 4: Calculate moles of H2O

Propane is the limiting reactant.

For 1 mol of propane consumed, we need 5 moles of O2 to produce 4 moles of H2O and 3 moles of CO2

For 0.726 moles of propane we'll have 4*0.726 = 2.904 moles of H2O

Step 5: Calculate mass of H2O

Mass of H2O = moles of H2O * molar mass of H2O

Mass of H2O = 2.904 moles * 18.02 g/mol

Mass of H2O = 52.33 grams

There is 52.33 grams of water produced.

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