Answer:
Explanation:
Firstly, we have to determine the mass of metal X. We can do that by interpreting the first and second statement mathematically.
Metal X can form 2 oxides (A and B).
A + B = 3g
The mass of oxygen in A is 0.72g and the mass of oxygen in B is 1.16g.
The mass of metal X in the two oxides will be the same because it's the same metal.
Thus, we represent the mass of the metal in the two oxides as 2X.
2X + 0.72 + 1.16 = 3
2X + 1.88 = 3
2X = 3 - 1.88
2X = 1.12
X = 0.56
<u>Thus, 0.56 g of the metal combines with 0.72g of oxygen in A and 1.16 g of oxygen in B.</u>
Thus, mass of metal (X) in 1g of oxygen in A is
0.56g ⇒ 0.72g
X ⇒ 1
X = 1 × 0.56/0.72
X = 0.78 g
Hence, 0.78g of the metal will combine with 1g of oxygen for A
Also, mass of metal (X) in 1g of oxygen in B is
0.56g ⇒ 1.16g
X ⇒ 1g
X = 1×0.56/1.16
X = 0.48 g
Thus, 0.48g of the metal will combine with 1g of oxygen for B
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Answer:
71.5g
Explanation:
The reaction equation is given as:
C + O₂ → CO₂
Mass of C = 42g
Mass of O₂ = 52g
Unknown:
Mass of CO₂ produced = ?
Solution
Now to solve this problem, we have to find limiting reactant which is the one given in short supply in this reaction.
The extent of the reaction is controlled by this reactant.
Find the number of moles of the given species;
Number of moles =
Number of moles of C =
= 3.5mol
Number of moles of O₂ =
= 1.63mol
Now;
From the balanced reaction equation;
1 mole of C reacted with 1 mole of O₂
We see that C is in excess and O₂ is the limiting reactant.
1 mole of O₂ will produce 1 mole of CO₂
So; 1.63mole of O₂ will produce 1.63 mole of CO₂
Mass of CO₂ = number of moles x molar mass
Molar mass of CO₂ = 44g/mol
Mass of CO₂ = 1.63 x 44 = 71.5g