After 1 half life its half the element
after 2 hl its quarter remaining
after 3 hl its half the quarter which is 1 eighth
after 3 half lifes
Answer:
Yes, yield.
Explanation:
N2(g) + 3 H2(g) → 2 NH3 (g) balanced equation
First, find limiting reactant:
Moles H2 = 1.83 g x 1 mole/2 g = 0.915 moles H2
Moles N2 = 9.84 g N2 x 1 mole/28 g = 0.351 moles N2
The mole ratio of H2: N2 is 3:1, so H2 is limiting (0.915 is less than 3 x 0.351)
Theoretical yield of NH3 = 0.915 mol H2 x 2 mol NH3/3 mol H2 = 0.61 moles NH3
Answer:- 448 mL of hydrogen gas are formed.
Solution:- It asks to calculate the volume of hydrogen gas formed in milliliters at STP when 0.020 moles of magnesium reacts with excess HCl acid. The balanced equation is:
There is 1:1 mol ratio between Mg and hydrogen gas. So, the moles of hydrogen gas is also equals to the moles of Mg reacted.
moles of Hydrogen gas formed = 0.020 mol
At STP, volume of 1 mol of the gas is 22.4 L. We need to calculate the volume of 0.02 moles of hydrogen gas.
= 0.448 L
They want answer in mL. So, let's convert L to mL using the conversion formula, 1L = 1000mL
= 448 mL
So, 0.020 moles of magnesium would produce 448 mL of hydrogen gas at STP on reacting with excess of HCl acid.
Answer:
HCl acts as an excess reagent
Explanation:
The reaction equation is;
PbO + 2HCl → PbCl2 + H20
Number of moles of PbO = 13g/223.2 g/mol = 0.058 moles
Since the mole ratio is 1:1, 0.058 moles of PbCl2 is produced
Number of moles of HCl = 6.4 g/36.5g/mol = 0.175 moles
2 moles of HCl yields 1 mole of PbCl2
0.175 moles yields 0.175 moles * 1/2
= 0.0875 moles of PbCl2
Hence; PbO yields the least number of moles of product so it is the limiting reactant and HCl is the reactant in excess