Answer:
83.20 g of Na3PO4
Explanation:
1 mole of Na3PO4 contains 3 moles of Na+.
Mole of Na ion to be prepared = Molarity x volume
= 0.700 x 725/1000
= 0.5075 mole
If 1 mole of Na3PO4 contains 3 moles of Na ion, then 0.5075 Na ion will be contained in:
0.5075/3 x 1 = 0.1692 mole of Na3PO4
mole of Na3PO4 = mass/molar mass = 0.1692
Hence, mass of Na3PO4 = 0.1692 x molar mass
= 0.1692 x 163.94
= 83.20 g.
83.20 g of Na3PO4 will be needed.
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Carbohydrates is the substance that makes up fats
Since
21.2 g H2O was produced, the amount of oxygen that reacted can be obtained
using stoichiometry. The balanced equation was given: 2H₂ + O₂ → 2H₂O and
the molar masses of the relevant species are also listed below. Thus, the
following equation is used to determine the amount of oxygen consumed.
Molar mass of H2O = 18
g/mol
Molar mass of O2 = 32
g/mol
21.2 g H20 x 1 mol
H2O/ 18 g H2O x 1 mol O2/ 2 mol H2O x 32 g O2/ 1 mol O2 = 18.8444 g O2
<span>We then determine that
18.84 g of O2 reacted to form 21.2 g H2O based on stoichiometry. It is
important to note that we do not need to consider the amount of H2 since we can
derive the amount of O2 from the product. Additionally, the amount of H2 is in
excess in the reaction.</span>
