a. If 50 mL of 0.2 mol/L of tin (IV) phosphate is mixed with an excess of sodium carbonate, how many grams of tin (IV) carbonate will form?

The mass of tin (IV) carbonate, Sn(CO₃)₂, formed when 50 mL of 0.2 mol/L of tin (IV) phosphate is mixed with excess sodium carbonate is 7.17 g

We'll begin by calculating the number of mole of Sn₃(PO₄)₄ in the solution. This can be obtained as follow:

Volume = 50 mL = 50 / 1000 = 0.05 L

Molarity of Sn₃(PO₄)₄ = 0.2 mol/L

Mole of Sn₃(PO₄)₄ =?

Mole = Molarity × Volume

Mole of Sn₃(PO₄)₄ = 0.2 × 0.05

Mole of Sn₃(PO₄)₄ = 0.01 mole

Next, we shall determine the number of mole of Sn(CO₃)₂ produced by the reaction of 0.01 mole of Sn₃(PO₄)₄. This can be obtained as follow:

Sn₃(PO₄)₄ + 6Na₂CO₃ → 3Sn(CO₃)₂ + 4Na₃PO₄

From the balanced equation above,

1 mole of Sn₃(PO₄)₄ reacted to produce 3 moles of Sn(CO₃)₂.

Therefore,

0.01 mole of Sn₃(PO₄)₄ will react to produce = 0.01 × 3 = 0.03 mole of Sn(CO₃)₂.

Finally, we shall determine the mass of 0.03 mole of Sn(CO₃)₂.

Mole of Sn(CO₃)₂ = 0.03 mole

Molar mass of Sn(CO₃)₂ = 119 + 2[12 + (16×3)] = 239 g/mol

Mass of Sn(CO₃)₂ =?

Mass = mole × molar mass

Mass of Sn(CO₃)₂ = 0.03 × 239

Mass of Sn(CO₃)₂ = 7.17 g

Therefore, the mass of tin (IV) carbonate, Sn(CO₃)₂, formed from the reaction is 7.17 g

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