Whew. long question.
Barium carbonate is the source of barium compounds. It is produced in an aqueous precipitation reaction from barium sulfide and sodium carbonate. (Barium sulfide is a soluble compound obtained by heating the mineral barite, which is barium sulfate, with carbon.) What are the molecular equation and net ionic equation for the precipitation reaction? A solution containing 33.9 g of barium sulfide requires 21.2 g of sodium carbonate to react completely with it, and 15.6 of sodium sulfide is produced in addition to whatever barium carbonate is obtained. How many grams of barium sulfide are required to produce 10.0 tons of barium carbonate?
BaS(aq)+Na2CO3(aq)->BaCO3(s)+Na2S(aq)
Ba+2(aq)+CO3-2(aq)->BaCO3(s)
7.78(with a line showing sig figs)41266 x 10^6 g BaS
I worked it and got 7.7876 x 10^6 which I would round to 7.79*10^6. I checked the conversions with an on-line calculator and it converted to 7.7875 x 10^6. I used 169.393 for the molar mass of BaS, 197.33 for the molar mass of BaCO3, 2000 lbs in a ton and 453.59 g in a pound.
To find the number of grams of barium sulfide required to produce 10.0 tons of barium carbonate, we need to use the given information about the stoichiometry of the reaction and perform a series of conversions.
Let's start by calculating the molar mass of barium sulfide (BaS). Barium (Ba) has a molar mass of 137.33 g/mol, and sulfur (S) has a molar mass of 32.07 g/mol. Therefore, the molar mass of BaS is:
Molar mass of BaS = (1 × Ba molar mass) + (1 × S molar mass)
= (1 × 137.33 g/mol) + (1 × 32.07 g/mol)
= 169.40 g/mol
Now, let's use the given information that 33.9 g of barium sulfide reacts with 21.2 g of sodium carbonate (Na2CO3) to find the stoichiometry of the reaction.
From the balanced equation: 1 mol BaS reacts with 1 mol Na2CO3
To convert grams to moles, we need to divide the given mass by the molar mass of the corresponding compound.
Moles of BaS = (Given mass of BaS) / (Molar mass of BaS)
= 33.9 g / 169.40 g/mol
= 0.20 mol BaS
Since one mole of BaS produces one mole of BaCO3, we can use this relationship to find the moles of BaCO3 produced.
Moles of BaCO3 = 0.20 mol BaS
To convert moles to grams, we multiply the moles by the molar mass of BaCO3.
Mass of BaCO3 = (Moles of BaCO3) × (Molar mass of BaCO3)
= 0.20 mol BaCO3 × (1 × (Ba molar mass) + 1 × (C molar mass) + 3 × (O molar mass))
= 0.20 mol BaCO3 × (1 × 137.33 g/mol + 1 × 12.01 g/mol + 3 × 16.00 g/mol)
= 0.20 mol BaCO3 × 197.34 g/mol
= 39.47 g BaCO3
So, for every 33.9 g of BaS, we obtain 39.47 g of BaCO3.
Now, to find the grams of BaS required to produce 10.0 tons of BaCO3, we continue with the following calculations:
Mass of BaS required = (Mass of BaCO3 required) × (Mass of BaS / Mass of BaCO3)
= (10.0 tons × 2000 lb/ton × 0.4536 kg/lb) × (39.47 g BaS / 3 9.47 g BaCO3)
≈ 7.78 × 4.12 × 10^6 g BaS (rounded to the appropriate number of significant figures)
Therefore, approximately 7.78 × 10^6 g of barium sulfide (BaS) are required to produce 10.0 tons of barium carbonate (BaCO3).
To find the number of grams of barium sulfide required to produce 10.0 tons of barium carbonate, we can use stoichiometry.
First, convert tons to grams:
10.0 tons = 10.0 x 1000 x 1000 grams = 10,000,000 grams
Now, we need to set up a ratio using the balanced equation to find the number of grams of barium sulfide. From the balanced equation, we know that for every 1 mole of BaCO3 produced, we need 1 mole of BaS. The molar mass of BaS is 169.39 g/mol.
So, the calculation is as follows:
10,000,000 grams BaCO3 x (1 mole BaCO3/197.34 grams BaCO3) x (1 mole BaS/1 mole BaCO3) x (169.39 grams BaS/1 mole BaS)
= 85,191,338 grams BaS
Therefore, approximately 85,191,340 grams of barium sulfide are required to produce 10.0 tons of barium carbonate.