286 (mol BaCl2) * 208.23 (g/mol BaCl2) = 59,553.78 = 59.6*103 g BaCl2 = 59.6 kg BaCl2
First, calculate the moles of HCl in the reaction using the volume and molarity provided. Since it is a 1:1 neutralization reaction, the moles of Ba(OH)2 are equal to the moles of HCl. Next, calculate the mass of barium chloride using the molar mass provided and the moles of BaCl2 produced in the reaction.
To convert moles to grams, multiply the number of moles by the molar mass of the compound. The molar mass of barium nitrite (Ba(NO2)2) can be calculated as: 1Ba + 2(NO2) = 137.33 + 2(46.01) = 229.35 g/mol. So, 4.50 moles of barium nitrite would be 4.50 moles x 229.35 g/mol = 1032.08 grams of barium nitrite.
To determine the number of moles in 0.98 grams of Potassium chloride, you need to divide the given mass by the molar mass of Potassium chloride. The molar mass of KCl is approximately 74.55 g/mol. So, 0.98 grams / 74.55 g/mol ≈ 0.013 moles of KCl.
There are two chloride ions in one formula unit of barium chloride.
To find the number of grams of ammonium chloride in 0.500L of a 2.00M solution, we need to use the formula: concentration (M) = moles of solute / volume of solution (L). First, calculate the moles of ammonium chloride present in 0.500L of the solution using C = n/V. Then, convert moles to grams by multiplying by the molar mass of ammonium chloride (NH4Cl).
Since barium sulfate and barium chloride have a 1:1 molar ratio, you would need the same amount of barium chloride as barium sulfate, so 100 grams.
89.3
Since both barium chloride and barium sulfate contain one mole of barium atoms pert mole of compound, the moles of barium sulfate will be the same, 0.100, when barium has the limiting concentration in the production of the sulfate.
3,75 moles barium chloride
23.3772 grams are there in four tenths moles of sodium chloride
First, you want to figure out how many grams 1 mol of barium is. Look at the periodic table of elements. (www.webelements.com) Barium is Ba. Scroll down a little bit after clicking and you will see atomic weight. It is 137g. Now get the atomic weight for chlorine. It is Cl. The atomic weight is 35.5g. Just add them together. 137+35.5 = 172.5g = 1 mol of barium chloride. For 188g, just divide it by 172.5g. Which is 1.09 moles.
First, calculate the moles of HCl in the reaction using the volume and molarity provided. Since it is a 1:1 neutralization reaction, the moles of Ba(OH)2 are equal to the moles of HCl. Next, calculate the mass of barium chloride using the molar mass provided and the moles of BaCl2 produced in the reaction.
0,40 moles of sodium chloride contain 23,376 g.
To convert moles to grams, multiply the number of moles by the molar mass of the compound. The molar mass of barium nitrite (Ba(NO2)2) can be calculated as: 1Ba + 2(NO2) = 137.33 + 2(46.01) = 229.35 g/mol. So, 4.50 moles of barium nitrite would be 4.50 moles x 229.35 g/mol = 1032.08 grams of barium nitrite.
To determine the number of moles present, we first need to find the molar mass of barium (Ba), which is approximately 137.33 g/mol. Next, we use the formula n = m/M, where n is the number of moles, m is the mass of the sample (22.3 grams), and M is the molar mass (137.33 g/mol). By substituting these values, we find that there are approximately 0.162 moles of barium in the sample.
9.11 g
10 moles of sodium chloride have 584,397 g.