Question

calculate the grams of barium sulfate produced if 5.00 grams of aqueous barium nitrate is mixed with 4.00 grams of aqueous magnesium sulfate. How much excess reagent remains?

Answer 1

1)

Molar mass of Ba(NO3)2 = 1*MM(Ba) + 2*MM(N) + 6*MM(O)

= 1*137.3 + 2*14.01 + 6*16.0

= 261.32 g/mol

mass of Ba(NO3)2 = 5.0 g

we have below equation to be used:

number of mol of Ba(NO3)2,

n = mass of Ba(NO3)2/molar mass of Ba(NO3)2

=(5.0 g)/(261.32 g/mol)

= 1.913*10^-2 mol

Molar mass of MgSO4 = 1*MM(Mg) + 1*MM(S) + 4*MM(O)

= 1*24.31 + 1*32.07 + 4*16.0

= 120.38 g/mol

mass of MgSO4 = 4.0 g

we have below equation to be used:

number of mol of MgSO4,

n = mass of MgSO4/molar mass of MgSO4

=(4.0 g)/(120.38 g/mol)

= 3.323*10^-2 mol

we have the Balanced chemical equation as:

Ba(NO3)2 + MgSO4 ---> BaSO4 + Mg(NO3)2

1 mol of Ba(NO3)2 reacts with 1 mol of MgSO4

for 0.0191 mol of Ba(NO3)2, 0.0191 mol of MgSO4 is required

But we have 0.0332 mol of MgSO4

so, Ba(NO3)2 is limiting reagent

we will use Ba(NO3)2 in further calculation

Molar mass of BaSO4 = 1*MM(Ba) + 1*MM(S) + 4*MM(O)

= 1*137.3 + 1*32.07 + 4*16.0

= 233.37 g/mol

From balanced chemical reaction, we see that

when 1 mol of Ba(NO3)2 reacts, 1 mol of BaSO4 is formed

mol of BaSO4 formed = (1/1)* moles of Ba(NO3)2

= (1/1)*0.0191

= 1.913*10^-2 mol

we have below equation to be used:

mass of BaSO4 = number of mol * molar mass

= 1.913*10^-2*2.334*10^2

= 4.47 g

Answer: 4.47 g of BaSO4 forms

2)

From balanced chemical reaction, we see that

when 1 mol of Ba(NO3)2 reacts, 1 mol of MgSO4 reacts

mol of MgSO4 reacted = (1/1)* moles of Ba(NO3)2

= (1/1)*0.0191

= 1.913*10^-2 mol

mol of MgSO4 remaining = mol initially present - mol reacted

mol of MgSO4 remaining = 0.0332 - 0.0191

mol of MgSO4 remaining = 0.0141 mol

Molar mass of MgSO4 = 1*MM(Mg) + 1*MM(S) + 4*MM(O)

= 1*24.31 + 1*32.07 + 4*16.0

= 120.38 g/mol

we have below equation to be used:

mass of MgSO4,

m = number of mol * molar mass

= 1.409*10^-2 mol * 120.38 g/mol

= 1.70 g

Answer:1.70 g of MgSO4 remains