What mass of NaHCO3 (MW= 84.01 g ) reacts in order to produce 250. g of...

Question

Question

What mass of NaHCO3 (MW= 84.01 g ) reacts in order to produce 250. g of Na3PO4 (MW= 163.94)

science chemistry

Answer 1

Answer #1

Answer:

The balanced equation between NaHCO₃ and Na₃PO₄ is

3 NaHCO₃ + H₃PO₄ = Na₃PO₄ + 3 H₂O + 3 CO₂

Given mass of Na₃PO₄=250 g and molar mass of Na₃PO₄=163.94 g/mol.

Therefore moles of Na₃PO₄=mass/molar mass=250 g/163.94 g/mol=1.524 mol

From the balanced equation, moles of NaHCO₃ required=3 x mol Na₃PO₄=3 x 1.524 mol=4.574 mol.

The molar mass of NaHCO₃=84.01 g/mol.

Therefore the mass of NaHCO₃ required= moles x molar mass=4.574 mol x 84.01 g/mol=384.33 g.

The mass of NaHCO₃ required = 384.33 g.

Please let me know if you have any doubt. Thank you.

Answer 2

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