Question

22.5 g of silver nitrate reacts with excess magnesium bromide, determine the mass

of magnesium nitrate that will be produced.

Answer 1

9.82 g of Mg(NO₃)₂

Step-by-step explanation:

Let's determine the reaction:

2AgNO₃ + MgBr₂ → Mg(NO₃)₂ + 2AgBr

2 moles of nitrate silver reacts with MgBr₂ in order to produce 1 mol of magnesium nitrate and silver bromide.

We determine the moles of AgNO₃

22.5 g . 1mol / 169.87g = 0.132 moles

Ratio is 2:1.

2 moles of silver nitrate can produce 1 mol of magnesium nitrate

Then, our 0.132 moles may produce (0.132 . 1)/ 2 = 0.0662 moles

We convert moles to mass:

0.0662 mol . 148.3 g/ mol = 9.82 g