How many grams of solid BaSO4 will form when Na2SO4 reacts with 25mL of 0.50 M Ba(NO3)2? _Ba(NO3)2 + _Na2SO4 ---> _ BaSO4 + _ NaNO3 Show all calculations

Md Ariful Islam

0 like 0 dislike
1 views
Share with your friends

1 Answers

Md Ariful Islam

Call
Answer: 1) Balance the reaction equation Ba(NO3)2 + Na2SO4 ---> BaSO4 + 2 NaNO3 2) Calculate number of moles of Ba(NO3)2 = 0.50 moles/L x 0.025 L = 0.0125 moles 3) Using stoichiometry calculate number of moles of BaSO4 =0.0125 moles 4) Calculate mass of BaSO4 = 0.0125 moles x 233.4 g/mole = 2.92 g BaSO4
0 like 0 dislike
1 views
Talk Doctor Online in Bissoy App
When 15.25 grams of A reacts with 34.22 grams of B, 18.69 grams of C is produced. How many grams of D are produced?

30.78 g

1 Answers 1 views
A 95.5 grams of sodium sulfate (Na₂SO₄)is dissolved in 1150 grams of distilled H2O. Calculate the molality of the Na₂SO₄ solution. Molar mass: Na = 23 g/mol ; S = 32 g/mol ; O = 16 g/mol

Molar mass of Na2SO4= 142.04 g/mol95.5/142.04= 0.6720.672mol/1.15kg= 0.585mol/Kg

1 Answers 1 views
3 BaCl2 + Fe2(SO4)3 ---> 3 BaSO4 + 2 FeCl3 If 25.0 grams of Barium Sulfate are produced during the reaction, how many grams of Iron(III) sulfate were used?

x 25.0 3 BaCl2 + Fe2(SO4)3 ---gt; 3 BaSO4 + 2 FeCl3 400 3* 233 x=400*25 / (3*233)=14.306 g

1 Answers 1 views
How much solid Pb(No3)2 must be added to 1L of 0.001M Na2So4 sloution for a pricipitation of PbSo4 (ksp=1.6*10^-8) to form ? assume no change in volume when solid is added

Pb(NO3)2 + Na2SO4 = 2NaNO3 + PbSO4; Pb2+ + 2NO3- + 2Na+ + SO42- = 2Na+ + 2NO3- + PbSO4 (s); Pb2+ + SO42- = PbSO4 (s). Ksp = [Pb2+]*[SO42-]...

1 Answers 1 views
For the reaction 2NaOH + H2SO4--> Na2SO4 + 2H2O, how many moles of Na2SO4 will be produced from 3.50 moles of NaOH if excess H2SO4 is available?

n(NaOH) = 3.50 molesSolution:The equation of a chemical reaction:2NaOH + H2SO4 --> Na2SO4 + 2H2OAccording to the chemical equation: n(NaOH)/2 = n(Na2SO4). n(Na2SO4) = n(NaOH)/2;n(Na2SO4) = 3.50 moles / 2...

1 Answers 1 views
as2s3+nano3+na2co3 -> na3aso4 + na2so4 + nano2 +co2 determine the theoretical mass of reactants needed to form 50 g of sodium arsenate

As2S3+14NaNO3+6Na2CO3 -> 2Na3AsO4 + 3Na2SO4 + 14NaNO2 +6CO2M(Na3AsO4)=208 g/moln (Na3AsO4)=m/M = 50/208 = 0.24 molM (As2S3) = 246 g/molM (NaNO3) = 85 g/molM (Na2CO3) = 106 g/moln (As2S3) = 1/2...

1 Answers 1 views
A student performed a precipitation reaction using 0.527 g of BaCl2 dissolved in water with an excess of Na2SO4. If 0.551 g of BaSO4 is recovered, what was the percent yield of BaSO4?

mp (BaSO4) = 0,551 g m (BaCl2) = 0,527 g η(BaSO4) - ? Na2SO4+ BaCl2 = 2NaCl + BaSO4↓ nt (BaSO4)= n (BaCl2) = m(BaCl2)/ M(BaCl2)= 0,527g/ 208g/mol = 0,003mol....

1 Answers 1 views
How many grams of sodium hydroxide would you need to produce 75 g of lead (II) hydroxide? 2 NaOH(aq) + Pb(NO3)2(aq) → 2 NaNO3(aq) + Pb(OH)2(s)

n(Pb(OH)2) = m/M = 75 g/241.21 g/mol = 0.31 molm(NaOH) = n*M = 2*n(Pb(OH)2)*M(NaOH) = 24.8 g

1 Answers 1 views
How many grams of barium sulfate can be made from 2.00 grams of barium chloride? BaCl2(aq) + Na2SO4(aq)→ BaSO4(s) + 2NaCl(aq)

Molar Mass of BaCl2 = 208.23 g/mol .Mole of BaCl2 = 2.00 grams of barium chloride / 208.23 g/mol ;So , Moles of BaCl2 = 0.0096 mole 1 mol. of BaCl2 gives...

1 Answers 1 views
When 25.00 mL of 3.42 M Pb(NO3)2 reacts with excess NaCl, how many grams of PbCl2 form? Pb(NO3)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2 NaNO3(aq)

Mole = concentration × volume Mole = 3.42M ×( 25 × 10-3 L)Mole = 0.0855 mole1 mole of Pb(NO3)2 = 278.1 g PbCl20.0855 mole = 278.1 × 0.0855 g PbCl2...

1 Answers 1 views
X