Explanation:
a.H2PO4 + 2KOH = K2PO4 + 2H2O
b. H2PO4 _ 2KOH
1(2) + 31 + 16(4) _ 2(23) + 2(16) + 2(1)
2 + 31 + 64 _ 46 + 32 + 2
97 _ 80
no of moles of H2PO4 = 2.5 ÷ 97
= 0.026mol
no of moles of 2KOH = 4 ÷ 80
= 0.05mol
2KOH is the limiting reactant
c. H2PO4 is the excess reactant
d. 1 mole H2PO4 weighs 2.5g
0.026mol H2PO4 weighs g
(0.026 × 2.5)
H2PO4 weighed 0.065g after reaction
Answer:
14.0067 g/mol
Explanation:
The molar mass of pure nitrogen is 14.0067 because when you look at the top right of an element on the periodic table of elements you can see the molar mass of an element. This can go for any other elements on the table, just look at the top right of the box. The molar mass of an element is also called the atomic weight. Hope this helps! :)
Answer:
46 g
Explanation:
The balanced equation of the reaction between O and NO is
2 NO + O₂ ⇔ 2 NO₂
Now, you need to find the limiting reagent. Find the moles of each reactant and divide the moles by the coefficient in the equation.
NO: (80 g)/(30.006 g/mol) = 2.666 mol
(2.666 mol)/2 = 1.333
O₂: (16 g)/(31.998 g/mol) = 0.500 mol
(0.500 mol)/1 = 0.500 mol
Since O₂ is smaller, this is the limiting reagent.
The amount of NO₂ produced will depend on the limiting reagent. You need to look at the equation to determine the ratio. For every mole of O₂ reacted, 2 moles of NO₂ are produced.
To find grams of NO₂ produced, multiply moles of O₂ by the ratio of NO₂ to O₂. Then, convert moles of NO₂ to find grams.
0.500 mol O₂ × (2 mol NO₂/1 mol O₂) = 1.000 mol NO₂
1.000 mol × 46.005 g/mol = 46.005 g
You will produce 46 g of NO₂.
Answer:
A more reactive halogen displaces a less reactive halogen from a solution of one of its salts.
Chlorine is more reactive than bromine, thus it displaces bromine.