From Lab #2 it was seen that the reaction between magnesium metal and hydrochlor

Question

From Lab #2 it was seen that the reaction between magnesium metal and hydrochloric acid produces magnesium chloride and hydrogen gas according lo the following equation Mg(S) + HCI (aq) + MgCl_2 (aq) + H_2(g) Balance the above equation. How many moles of hydrogen gas will form If 0.1234 grams of magnesium are used? How many grams of magnesium are required to produce 5 00 moles of magnesium chloride How many grams of magnesium chloride will form If 0.160 moles of hydrochloric acid are used? Sodium iodide reacts with lead (II) nitrate to form lead (II) iodide solid and sodium nitrate solution according to the following equation : Given 1 38 grams of Nal, how many grams of each product can be made

Explanation / Answer

7a

Mg + HCl = MgCl2 + H2

balance:

Mg + 2HCl = MgCl2 + H2

7b

m = 0.1234 g of Mg are used; mol of H2?

mol of Mg = mass/MW = 0.1234/24.305 = 0.005077 mol of Mg

then ratio is 1 mol of Mg per mol of H2 ; then 0.005077 mol of H2 are produced

7c

Mg required for 5 mol of MgCl2?

ratio is 1:1 according to reaction so

we need 5:5; 5 mol of Mg to produce 5 mol of MgCl

7d

g of MGCl if 0.16 mol ofHcl is used

2 mol of HCl produces : 1 mol of MgCl2

then

0.16 mol of Hcl --> 0.16/2 = 0.08 mol of MgCl2

mass = mol*MW = 0.08*95.211 = 7.61688 g of Mgcl2 produced

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