If 20.0g of magnesium react with excess hydrochloric acid, how many grams of mag

Question

If 20.0g of magnesium react with excess hydrochloric acid, how many grams of magnesium chloride are produced? Hydrogen gas is also produced in this reaction. Mg + HCl rightarrow 3gcl + H_2 How many grams of chlorine gas must be reacted with excess sodium iodide if 10.Og of sodium chloride are needed? Solid iodine is also produced in this single displacement reaction. Cl_2 + naI rightarrow nacl + I_2 How many grams of oxygen are produced in the decomposition of 5.00g of potassium chlorate? Potassium chloride is also produced in this reaction. KclO_3 rightarrow O_2 + Kcl What mass of copper(ll) is required to replace silver from 4.00g of silver (I) nitrate? Cu + AgNo_3 rightarrow cu (n O_3)_2 + Ag If excess ammonium sulfate reacts with 20.0g of calcium hydroxide to produce calcium sulfate, ammonia, and water, how many grams of ammonia (NH_3) are produced? If excess sulfuric acid reacts with 30.0g of sodium chloride, how many grams of HCI are produced? H_2SO_4 + Nacl rightarrow Hcl + (na_2)SO_4 How much silver (I) phosphate is produced if 10.0g of silver (I) acetate react with excess sodium phosphate in the unbalanced equation: AgCH_3COO + Na_3PO_4 rightarrow Ag_3 PO_4 + NaCH_3 COO How many grams of sodium hydroxide are needed to completely neutralize 25.0g of sulfuric acid to form sodium sulfate and water? NaOH + H_2 SO_4 rightarrow Na_2 SO_4 + H_2 O

Explanation / Answer

1. Mg + 2HCl --> MgCl2 + H2

1 mole of Mg forms 1 mole of MgCl2 i.e. 24.3 g of Mg forms (24.3 + 35.5 * 2) g = 95.3 g of MgCl2

Hence, 20 g of Mg forms 95.3 * 20/24.3 g = 78.44 g of MgCl2

2. Cl2 + 2NaI ---> 2NaCl + I2

2 * (23 + 35.5) g = 117g of NaCl is obtained from 35.5 * 2 g = 71 g of Cl2

Hence, 10 g of NaCl is obtained from 71/117 * 10 g = 6.07 g of Cl2

3. 2 KClO3 ---> 2KCl + 3O2

2 moles of KClO3 = 2 * (39 + 35.5 + 16 *3)g = 245 g KClO3 produces 3 * 16 * 2 g = 96 g of O2

Hence, 5.00 g KClO3 produces 96/245 * 5 g = 1.96 g O2

4. Cu + 2 AgNO3 ---> Cu(NO3)2 + 2 Ag

2 * (107.8 + 14 + 16 *3) g = 339.6 g of AgNO3 requires 63.55 g of Cu

Hence, 4.00 g of AgNO3 requires 63.55 * 4.00/339.6 g = 0.749 g of Cu

Post rest of the questions as a separate set- as per regulations of Chegg.

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