Final answer:
To find the limiting reactant for the given chemical equation, first balance the reaction, then calculate the moles of each reactant and determine which produces the least amount of product. The mass of the product can be found by converting the reacted moles of the limiting reactant into moles of the product and then to grams.
Step-by-step explanation:
Limiting Reactant and Product Mass Calculation
To solve this chemistry problem, we need to balance the reaction and calculate the limiting reactant and the amount of product formed. Starting with the given unbalanced chemical equation:
H3PO3 + Mg → Mg3(PO3)2 + H2
First, we balance the equation:
2H3PO3 + 3Mg → Mg3(PO3)2 + 3H2
Given the masses of reactants, 21.75 g of H3PO3 and 15.00 g of Mg, to identify the limiting reactant, we must convert the mass of each reactant to moles, use the balanced chemical equation to determine the moles of product each would produce, and then compare these amounts:
- Molar mass of H3PO3: approximately 81.99 g/mol
- Molar mass of Mg: approximately 24.31 g/mol
- Moles of H3PO3: 21.75 g / 81.99 g/mol
- Moles of Mg: 15.00 g / 24.31 g/mol
Using the stoichiometric coefficients from the balanced equation, calculate the moles of Mg3(PO3)2 each reactant can produce and then determine the limiting reactant by the one that produces the lesser amount of product. The mass of Mg3(PO3)2 can be calculated by converting the moles of the limiting reactant that reacted into moles of Mg3(PO3)2 and then to grams using its molar mass.