Asked by Loretta
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
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
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