Question

6. (16 points) Magnesium bromide reacts with silver nitrate to form silver bromide (solid) and magnesium sulfate (aq). A mixture of 50.0 g of magnesium bromide (MM - 184.1 g/mol) and 100 g of silver nitrate (MM- 169.9 g/mol) is allowed to react. Identify the limiting reactant and determine the mass of the excess reactant present in the vessel when the reaction is complete, Do by ICE Method Type of reactionDouble displacement 2115 MgBra (s) 2 AgNO (aq) > Mg(NO3)2(aq) + 2 Ag Br(s)

Answer 1

First step is to calculate the number of moles of magnesium bromide and silver nitrate.

$Number of moles= \frac{mass in gram}{molar mass}$

For magnesium bromide,

$Number of moles= \frac{50g}{184.1g/mol}=0.271moles.$

For silver nitrate,

$Number of moles= \frac{100g}{169.9g/mol}=0.588moles.$

Limiting reagent is defined as that reagent which is completely reacted during a reaction. Thus during a reaction, the limiting reagent will be the one which is present in smaller amount.

Here since the number of moles of magnesium bromide is less than that of silver nitrate, the limiting reagent will be magnesium bromide.

An excess reagent is the one which is present in excess after the formation of the product. Here the excess reagent is silver nitrate.

This is expressed using the ICE table,

	MgBr2(s)	+	2AgNO3(aq)	$\rightarrow$	Mg(NO3)2(aq)	+	2AgBr(s)
I	0.271 moles		0.588 moles		0 moles		0 moles
C	-0.271 moles		-2$\times$0.271 moles		0 moles		0 moles
E	0 moles		0.046 moles		0.271 moles		0.542 moles

Thus the moles of excess reactant present in the vessel after the completion of the reaction is 0.046 moles of silver nitrate.

Mass in gram is calculated as,

$Mass in gram= Molar mass \times Number of moles$

$=169.9g/mol\times0.046 moles=7.81g$