Molar mass:
O2 = 16 x 2 = 32.0 g/mol Mg = 24 g/mol
<span>2 Mg(s) + O2(g) --->2 MgO(s)
</span>
2 x 24.0 g Mg -------------> 32 g O2
5.00 g Mg -----------------> ( mass of O2)
mass of O2 = 5.00 x 32 / 2 x 24.0
mass of O2 = 160 / 48
= 3.33 g of O2
hope this helps!
Answer:
148.2 g of H20
Explanation:
Equation of reaction: 4NH3 + 5O2 ---> 4NO + 6H20
From the equation above, 4 moles of ammonia reacts with 5 moles of oxygen gas to produce 6 moles water.
Molar mass of NH3 = 17 g/mol;
Molar mass of O2= 32 g/mol;
Molar mass of H2O = 18 g/mol
First, we determine the limiting reactant:
4*17 g of NH3 reacts with 5*32 g of O2
Mass Ratio = 68 : 160
Therefore, NH3 is the limiting reactant.
68 g of NH3 reacts to produce 6* 18 g of H20 = 108 g of H2O
93.3 g of NH3 will react to produce (93.3 * 108)/68 g of H20 = 148.2 g of H2O
Therefore, the maximum amount of H2O produced = 148.2 g
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