How much sr(oh)2 • 8 h2o (m = 265.76) is needed to prepare 250.0 ml of solution in which [oh]= 0.100m?

The answer is = 3.322grams

The explanation:

1-First we need o get the no.of moles:

moles = molarity * volume(L)

by substitution:

∴ moles = 0.1M * 0.25 L

= 0.025 moles

but we have 2 OH in Sr(OH)2

so moles = 0.025 /2 = 0.0125 moles

2- then we need to get how many grams of it so, we are going to use this formula:

mass = moles * molar mass

by substitution:

∴ mass = 0.0125 * 265.76

= 3.322 grams

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