How many grams of chlorine gas must react to give 4.82g of BiCl3?
2Bi(s)+3Cl2(g)→2BiCl3(s)
My answer was wrong/please review and tell me where I went wrong.
4.82g BiCl3*1mol BiCl3/215.33g BiCl3*3 mol Cl2/2 mol Bicl3* 70.9g Cl2/1 mol Cl2=6.50 g Cl2
I just calculated it wrong.
My answer should have been 1.63g of Cl2
I think you problem is the molar mass of BiCl3. I think that is 315.34
The rest of it looks ok, at least at first glance.
1.63g of Cl2
To determine the correct answer, let's go through the calculations step-by-step:
1. Start with the given mass of BiCl3: 4.82g BiCl3.
2. Convert the mass of BiCl3 to moles using the molar mass of BiCl3. The molar mass of BiCl3 is 215.33g/mol, so:
4.82g BiCl3 * 1 mol BiCl3 / 215.33g BiCl3 = 0.0224 mol BiCl3.
3. Use the balanced chemical equation to find the mole ratio between BiCl3 and Cl2. From the balanced equation, we know that 2 moles of BiCl3 react with 3 moles of Cl2. Therefore:
0.0224 mol BiCl3 * 3 mol Cl2 / 2 mol BiCl3 = 0.0336 mol Cl2.
4. Finally, convert the moles of Cl2 back to grams using the molar mass of Cl2, which is 70.9g/mol:
0.0336 mol Cl2 * 70.9 g Cl2 / 1 mol Cl2 = 2.38g Cl2.
So, the correct answer is 2.38 grams of chlorine gas (Cl2) must react to give 4.82 grams of BiCl3.