Answer:
9.82 g of Mg(NO₃)₂
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
