How many grams of magnesium nitrate can be produced from 50.0 mL of a 2.0 mol/L nitric acid solution reacting with excess magnesium carbonate?
first write the balanced equation
MgCO3 (aq) + 2HNO3 (aq) ----> Mg(NO3)2 (aq) + H2O (l) + CO2 (g)
from th ebalanced equation it is clear hat for one mole of Mg(NO3)2 2 moles of HNO3 required
he said excess magnisium carbo nate so dont bauther about that
no of moles of HNO3 = molarity of HNO3 x volume of HNO3 in liters
= 2.0 mol/L x 0.05 L
= 0.1 mol
from 0.1 mol of HNO3 0.1/2 = 0.05 moles of Mg(NO3)2 will fom
grams of Mg(NO3)2 = moles of Mg(NO3)2 x molar mass of Mg(NO3)2
= 0.05 mol x 86.31 g/mol
= 4.32 grams
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