Answer:
58.0 g of MgO
Explanation:
in a perfect world, 70 g, however we don't live in a perfect world
The equation of reaction
2Mg + O₂ --> 2MgO
first find which element is limiting:
35 g x 1 mol/24.3 g of Mg x 2 mol of MgO/ 2 mole of Mg = 1.44 moles of MgO
35 g x 1 mol/32g of Mg x 2 mol of MgO/ 1 mole of O₂ = 2.1875 moles of MgO
This means Mg is the limiting factor, so you will be using this moles to find grams of MgO
1.44 mols of MgO x 40.3 g of MgO/ 1 mol = 58.0 g of MgO
The answer is B hope this helped
G(2)=2
For this, you can plug in 2 everywhere you see an n. So the equation will read:
g(2)=g(2-1)+2 -> g(2)=g(1)+2. Since we are given g(1)=0, we can plug in 0 where we see g(1). The equation is now. g(2)=0+2. So, g(2)=2.
Unfortunately, you failed to include the table 1 from which the molar heat capacity of aluminum could have been obtained. However, as a general rule, the heat needed to raise the temperature of a certain substance by certain degrees is calculated through the equation,
H = mcpdT
where H is heat, m is mass, cp is specific heat capacity, and dT is change in temperature. From a reliable source, cp for aluminum is equal to 0.215 cal/g°C. Substituting this to the equation,
H = (260.5 g)(0.215 cal/g°C)(125°C - 0)
H = 7000.94 cal
