How many grams of MgCl² are required to completely react with 500g of K³PO⁴?​

Sagot :

Solution:

Step 1: Write the balanced chemical equation.

3MgCl₂ + 2K₃PO₄ → Mg₃(PO₄)₂ + 6KCl

Step 2: Calculate the molar mass of substances needed.

For MgCl₂

molar mass = (24.31 g/mol × 1) + (35.45 g/mol × 2)

molar mass = 95.21 g/mol

For K₃PO₄

molar mass = (39.10 g/mol × 3) + (30.97 g/mol × 1) + (16.00 g/mol × 4)

molar mass = 212.3 g/mol

Step 3: Determine the mole ratio needed.

mole ratio = 3 mol MgCl₂ : 2 mol K₃PO₄

Step 4: Calculate the mass of MgCl₂ required using the mole ratio.

[tex]\text{mass of MgCl₂ = 500 g K₃PO₄} × \frac{\text{1 mol K₃PO₄}}{\text{212.3 g K₃PO₄}} × \frac{\text{3 mol MgCl₂}}{\text{2 mol K₃PO₄}} × \frac{\text{95.21 g MgCl₂}}{\text{1 mol MgCl₂}}[/tex]

[tex]\boxed{\text{mass of MgCl₂ = 336 g}}[/tex]

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