2 points If 4.42 g of aluminum are actually formed in this reaction, what is the
percent yield?
How many grams of aluminum will theoretically be produced from the reaction of
12.5 g of calcium and 47.3 g of aluminum bromide?
3Ca
+
2AlBrı →
3CaBr
+
2Al
127%
O O
78.8%
108%
0 0
92.3

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

(b) which contains higher levels of vitamin b, oranges or spinach?

explanation:

you can eliminate a, c, and d because they are questions based on opinions.