<span>LiOH+HBr---> LiBr +h20. Moles of LiOH = 10/24 = 0.41moles. According to stoichiometry, moles of LiOH = moles of LiBr = 0.41moles. Therefore mass of LiBr =moles of LiBr x molecular weight of LiBr = o.41 x 87 = 35.67g. Hope it helps </span>
Answer:
Calculate moles KCl: 2.55 g / 74.55 g/mol = 0.0342 moles KCl
Volume KCl = 0.0342 mol X (1 L/0.150 mol) = 0.228 L X 1000 mL/L = 228 mL
Explanation:
Answer:
See explaination
Explanation:
please kindly see attachment for the step by step solution of the given problem
Answer: Beryllium phosphide
Explanation:it has a molecular weigt of 88.98407 g/mol
its CID is 16453515
First, write the balanced equation:
2KClO3−→−heat2KCl+3O22KClO3→heat2KCl+3O2
Get the MrMr of KClO3KClO3 and KClKCl(122.55 g mol−1122.55 g mol−1 and 74.6 g mol−174.6 g mol−1 respectively)
Next, find out how many moles of KClO3KClO3 are present. From calculations (n=m/Arn=m/Ar), there are 0.463 mol of KClO3KClO3 present.
From the above, we can infer that 2 mol of KClO3KClO3 decompose to give 2 mol of KClKCl. So, by the complete thermal decomposition of 0.463 mol of KClO3KClO3, we get 0.463 mol of KClKCl.
0.463 mol of KClKCl is equal to 34.5 g of KClKCl.
Hence,
34.5 g of KClKCl was produced.
