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2 years ago

How many moles of sodium hydrogen carbonate are present in 2.61 grams of this compound?

Chemistry

1 answer:

Sloan [31]2 years ago

7 0

Molar mass of NaHCO_3

23+1+12+3(16)
36+48
84g/mol

Now

$\\ \sf\longrightarrow No\:of\:moles=\dfrac{Given\:mass}{Molar\:mass}$

$\\ \sf\longrightarrow No\:of\; moles=\dfrac{2.61}{84}$

$\\ \sf\longrightarrow No\:of\:moles=0.0312mol$

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Answer:

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Step 1: Data given

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Step 2: The balanced equation:

2 Al(s) + 2 NaOH(aq) + 6 H2O(l) → 2 NaAl(OH)4(aq) + 3 H2(g)

Step 3: Calculate moles of Al

Moles Al = mass Al / Molar mass Al

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Moles Al = 0.0678 moles

Step 4: Calculate moles of NaOH

Moles NaOH = 4.30 grams / 40 g/mol

Moles NaOH = 0.1075 moles

Step 5: Calculate limiting reactant

For 2 moles of Al, we need 2 moles of NaOH

Aluminium is the limiting reactant. It will completely be consumed ( 0.0678 moles)

NaOH is in excess. There will react 0.0678 moles

There will remain 0.1075 - 0.0678 = 0.0397 moles

Step 6: Calculate moles of hydrogen

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For 0.0678 moles of Al, there is produced 0.0678 *3/2 = 0.1017 moles of H2

Step 7: Calculate mass of H2

Mass of H2 = Moles H2 * Molar mass of H2

Mass of H2 = 0.1017 moles * 2.02 g/mol

Mass of H2 = 0.205 grams

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