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Answer:
look up chemical board it will help you
Explanation:
Using the ideal gas equation:
PV = nRT
Substituting n with mass / Mr
PV = mRT/Mr
Density = m/V
So rearranging:
Density = PMr/RT
P = 1 atm
R = 0.082 L atm / K mol
T = 273 K
Density = (1 x 80.6) / (0.082 x 273)
Density = 3.6 g / L
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