Hydrobromic acid dissolves solid iron according to the reaction:
Fe(s) + 2 HBr(aq) → FeBr2(aq) + H2( g)
What mass of HBr (in g) do you need to dissolve a 3.2-g pure iron bar on a padlock? What mass of H2 would the complete reaction of the iron bar produce?
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Hello. In this primary told the process of cad welding involves copper to oxide and aluminum reacting to form copper metal and aluminum oxide brass. To calculate the mass of aluminum needed to react with 6.4 g of copper to oxide and the massive copper that is formed assuming the reaction goes to completion. So we begin with our mass of copper to oxide. In order to relate anything in the balanced reaction equation, it has to be on a moral basis. So we'll convert our mass of copper to oxide to molds using a smaller mass. And then we can make use of the balanced reaction equation. We find that we have three moles of copper, two oxide reacts with two moles of aluminum based on the coefficients and the balanced reaction equation. And we're going to convert to the massive aluminum making use of its smaller mass. And so our grams of copper to oxide cancels moles of copper to oxide cancels moles of aluminum cancels. And we're left with a mass of Aluminum, which works out to 1.4 g of aluminum. And then we will again begin with our mass of copper to oxide to determine the mass of copper that is formed again, to relate copper to oxide to copper needs to be on the mole basis. So we'll convert our mass of copper to oxide two moles. And then now that we're moles, we can relate the moles of copper to oxide to that of copper. Three moles of copper to oxide then reacts to form three moles of copper. Now to find the massive copper will convert from moles to mass using a smaller mass. So our grams of copper to oxide cancels moles of copper dioxide cancels moles of copper cancels, and we're left with then 5.1 g of copper, so the amount of aluminum needed to react to 6.4 g of copper to oxide is 1.4 and the mass of copper that will be produced is 5. g. Thanks for watching Hope. This helped.