Final answer:
To find the maximum mass of chromium(III) phosphate produced, convert the mass of chromium and phosphoric acid to moles, determine the limiting reactant using the balanced equation, and then calculate the mass of the product based on the moles of the limiting reactant.
Step-by-step explanation:
The question is asking how much chromium(III) phosphate can be produced when 13.88 g of chromium metal reacts with 10.77 g of phosphoric acid. To answer this, we must first convert the mass of chromium and phosphoric acid to moles. Using the molar mass of chromium (51.996 g/mol) and phosphoric acid (H3PO4, 97.994 g/mol), calculate the number of moles of each reactant.
For chromium: Moles of Cr = 13.88 g / 51.996 g/mol
For phosphoric acid: Moles of H3PO4 = 10.77 g / 97.994 g/mol
Next, we need to look at the balanced equation for the reaction between chromium and phosphoric acid to produce chromium(III) phosphate (CrPO4). The stoichiometry is:
- 6 Cr + 2 H3PO4 -> 2 Cr3(PO4)2 + 6 H2
We can see from the equation that 6 moles of chromium are required to react with 2 moles of phosphoric acid to form 2 moles of chromium(III) phosphate. Using the mole ratio, find whether chromium or phosphoric acid is the limiting reagent and therefore determine how many moles of chromium(III) phosphate can form.
Once we have the moles of the product, we can calculate its mass by multiplying by its molar mass. Since chromium is often the limiting reagent in reactions, it's likely that it will determine the amount of product formed in this particular reaction.