Answer :
The 26.6 grams of copper(II) nitrate are needed to produce 39.4 grams of copper(II) phosphate in the presence of excess sodium phosphate.
Why the 26.6 grams of copper(II) nitrate are needed to prodouce 39.4 grams?
The balanced chemical equation for the reaction between copper(II) nitrate and sodium phosphate to form copper(II) phosphate is:
Cu(NO3)2 + Na3PO4 → Cu3(PO4)2 + 6NaNO3
From the equation, we can see that one mole of copper(II) nitrate reacts with one mole of sodium phosphate to produce one mole of copper(II) phosphate.
To determine the amount of copper(II) nitrate needed to produce 39.4 g of copper(II) phosphate, we need to use stoichiometry and the molar mass of copper(II) nitrate.
- Calculate the molar mass of copper(II) nitrate
- Cu(NO3)2 = 1 x Cu + 2 x N + 6 x O
- Cu(NO3)2 = 1 x 63.55 g/mol + 2 x 14.01 g/mol + 6 x 16.00 g/mol
- Cu(NO3)2 = 187.55 g/mol
- Calculate the moles of copper(II) phosphate produced
- m(Cu3(PO4)2) = 39.4 g ÷ (3 x 31.00 g/mol + 2 x 94.97 g/mol)
- m(Cu3(PO4)2) = 39.4 g ÷ 277.94 g/mol
- m(Cu3(PO4)2) = 0.1418 mol
- Use stoichiometry to determine the moles of copper(II) nitrate needed
- From the balanced chemical equation, we know that one mole of copper(II) nitrate reacts with one mole of copper(II) phosphate. Therefore, the moles of copper(II) nitrate needed is equal to the moles of copper(II) phosphate produced.
- n(Cu(NO3)2) = 0.1418 mol
- Calculate the mass of copper(II) nitrate needed
- m(Cu(NO3)2) = n(Cu(NO3)2) x M(Cu(NO3)2)
- m(Cu(NO3)2) = 0.1418 mol x 187.55 g/mol
- m(Cu(NO3)2) = 26.6 g
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