Calculate the volume (in mL) of NO2 at 1 atm and 273 K obtained by treating 12.7 g of Cu with an excess of concentrated nitric acid from the following reaction: Cu(s) + HNO3(aq) —• Cu(NO3)2(aq) + H200(l) + NO2

1). 8960 2 ). 896

3 ). 89.3 4). 6900

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explanation:

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as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2