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Sagot :
There are 372 grams of calcium phosphate are theoretically produced if we start with 3. 40 moles of calcium nitrate and 2. 40 moles of lithium phosphate .
Calculation,
The reactant or reagent that produces the least moles of the products is called limiting reagents. When limiting reagents used up , the reaction stops.
The balanced equation is given as,
[tex]3Ca(NO_{3} )_{2} + 2Li_{3} ( PO_{4} )[/tex] → [tex]3LiNO_{3} + Ca_{3}( PO_{4} )_{2}[/tex]
Multiply the moles of each reactant by the mole ratio between it and calcium phosphate in the balanced equation . so that the moles of the reactant cancel , leaving moles of calcium phosphate.
3.4 mol of calcium nitrate × 1 mol calcium phosphate / 3 mol calcium nitrate = 1.13 mol calcium phosphate
2.4 mol of lithium phosphate× 1 mol calcium phosphate / 2 mol lithium phosphate = 1.02 mol calcium phosphate
So, calcium nitrate is limiting reactant .
Calculation of mass of 1.02 mol calcium phosphate.
Multiply the moles of calcium phosphate by its molar mass.
molar mass of calcium phosphate = 3×40.078 g/mol calcium ion+2×30.9 g/mol phosphorus + 8×15.99 g/mol calcium phosphate = 310.178 g/mol calcium phosphate
1.20 mol calcium phosphate × 310.178 g/mol = 372 gram
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