Question

please solve and show work

Equivalent Weight of an Unknown Metal Experimental Set Up DISCUSSION: Consider the balanced equations for the following reactions of metals with HCl, each written for the production of į mole of H2 (one equivalent of H2) Na + HCI --> NaCl + H2 Ź Mg + HCI -> į MgCl2 + H2 į Al + HCI -> AlCl3 + H2 a. What additional datum would be needed in order to find the atomic mass of your metal from the equivalent mass? IDENTIFY your unknown metal from the equivalent mass and the Periodic Table You may encounter some difficulty, if you have an alloy. Show all setups. Be careful with significant figures b. b. The equation for the reaction of silver with dilute nitric acid is: 3 Ag + 4HNO3 -- 3AgNO3 + NO(g) + 2 H20. If 2.30 g of silver are dissolved in xs (excess) nitric acid, this is moles of silver, and you will liberate moles of NO gas, which will occupy liters at 60°C and 715 torr.

Answer 1

* for the equation of of Nitric so aid is 6 reaction - 3 Ag + 4HINO - 3 Ag Noz & NO(g) + 2tho (M. W. a 107 g/nol) No. of moles of silves & weight = 2.30 Mol weight lo7 = 0.021 moles since, 3 moles of Ag produces, moles of No = I moto. I - .. 0.021 " u n ". " WE 0.021 - moles of No produced a 0.00716 moles ces 7. 16 x 16 h moles. The conditions are at 60°c, and 716 torr - Using Ideal gas equation, PV = nRT [R = gas constant - VE URT [RE 62.36 L-torr/mol-k] too, Volume will be V= 0.00716 x 62.36 *333. [T = 333k ] 715 Tv = 0.208 Litres I is occupied