How many grams of ammonia are present in 5.0 L of a 0.050 M solution

4.25g

Explanation:

Molar concentration or molarity can be calculated thus;

Molarity (M) = number of moles (n) ÷ volume (V)

According to the provided information in this question, Volume = 5.0L, M = 0.050 M

number of moles = molarity × volume

n = 0.050 × 5

n = 0.25moles

Number of moles in a substance = mass (M) ÷ molar mass (MM)

Molar mass of ammonia (NH3) = 14 + 1(3)

= 17g/mol

Mass = moles × molar mass

Mass (g) = 0.25 × 17

Mass = 4.25g

i just calculated it all out, and not including weekends, and all the different days off. as of this school year (2018-2019) students are spending 2198 hours in school this year

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2