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How many grams of oxygen will react completely with a block of calcium metal that is 3.0 cm by 3.5 cm by 4.2 cm, if the density of calcium is 1.55 g/mL? Show all steps of your calculations as well as the final answer.
Ca + O2 ---> CaO
PLEASE HELP ITS A RESPONSE QUESTION

Respuesta :

(3.0 cm) (3.5 cm) (4.2 cm) = 44.1 cm3
aka 44.1 ml

find mass
44.1 ml @ 1.55 g/mL = 68.355 grams of Ca

the equation
Ca + O2 → CaO
relates that, using molar masses
1 mol of Ca @ 40.08 grams / mol... reacts with 1 mole of O2 @ 32.00 g/mol

so How many grams of oxygen gas will react completely with 68.355 grams of Ca
68.355 grams of Ca @ 32.00 g O2 / 40.08 g Ca = 54.68 g O2

you might be expected to round your answer to 2 sig figs
=since your data has 2 sig figs
if so your answer would be
55 grams of O2

Source of Information: (https://answers.yahoo.com/question/index?qid=20110811123311AAJwHiM)

JoshEast
Volume of Ca =  3.0 cm × 3.5 cm × 4.2 cm
                        = 44.1 cm³  = 44.1 mL


          Mass = volume × density
Mass of Ca = 44.1 mL × 1.55 g/mL
                    = 68.355 g


Now, since moles =  mass ÷ molar mass
 then moles of Ca = 68.355 g ÷ 40 g/mol
                               = 1.7089 mol


The balanced equation for the reaction between Ca and O₂ :
                        2 Ca   +   O₂  →  2 CaO


     the mole ratio of Ca  :  O₂  is  2  :  1
(which means that there are two times more moles of Ca than moles of O₂)

∴ if moles of Ca = 1.7089 mol

then moles of O₂ = 1.7089 mol  ÷ 2
                              =  0.8545 mol


Now,      mass  = moles × molar mass

      ⇒  mass of O₂ = 0.8545 mol × 32 g/mol
                              = 27.34 g

∴ the grams of oxygen will react completely with a block of calcium metal is 27.34 g

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