<span> Mg(OH)2(s) + 2HCl(aq) yield MgCl2(aq) + 2H2O(l)
grams HCl required = (50.6 grams Mg(OH)2) * (1 mol Mg(OH)2 / 58.3197 grams Mg(OH)2) * (2 mol HCl / 1 mol Mg(OH)2) * (36.453 grams HCl / 1 mol HCl) = 63.26 grams HCl required
Since there are only 45.0 grams HCl, then HCl is the limiting reactant.
theoretical yield MgCl2 = (45.0 grams HCl) * (1 mol HCl / 36.453 grams HCl) * (1 mol MgCl2 / 2 mol HCl) * (95.211 grams MgCl2 / 1 mol MgCl2) = 58.6 grams MgCl2 </span>
Answer
2-methyl-2-pentene
Explanation:
1. Identify the group that takes precedence in this case alkene hence this molecule is an alkene with a methyl group side chain.
2.Find the longest carbon chain where the functional group(alkene group in this case) has the lowest Carbon number
3.What are the side groups? One side group can be seen at carbon 2 this group is methyl
4. Naming, number separated by "," and number from letters by "-" so the compound should be
2-methyl-2-pentene
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You have 3 (h2(so4)) on the reactants side so you need to have 6 total hydrogen’s on the products side. Therefore 3(h2) is required.
Answer: I think it's a don't blame me if it's wrong though
Explanation:
