Question

1.) For the chemical reaction 2 Al + 3 H 2 SO 4 ⟶ 3 H 2 + Al 2 ( SO 4 ) 3 what mass of hydrogen is produced from 1.97 mol of aluminum?

2.) Convert 1.77×1024 atoms of carbon to moles of carbon.

1.77×1024 atoms=

3.) For the reaction

2KI+Pb(NO3)2⟶PbI2+2KNO3how many grams of lead(II) iodide, PbI₂, are produced from 89.7 g of potassium iodide, KI?

4.)For the reaction

2NaOH+H2SO4⟶Na2SO4+2H2O

how many grams of sulfuric acid, H₂SO₄, are needed to react completely with 33.1 g of sodium hydroxide, NaOH?

PLEASE HELP...I WOULD GREATLY APPRECIATE IT...

Answer 1

1) 2 Al + 3 H2SO4 ⟶ 3 H2 + Al2(SO4)3  this is balanced reaction equation.

That is 2 moles of Al (MW= 26.98) gives 3 moles of H2 (MW= 2)

Hence 1.97 mols of Al produced = 1.97 * 3 / 2 moles of Hydrogens  

= 2.955 moles of H2 produced

Hence 2.955 moles * mol wt. of H2  

= 5.91 g of Hydrogen are produced.

2) 1 mol = 6.022 X 1023 atoms ( Avogadro's number)

1.77 X 1024 atoms of carbon = 1.77 X 1024  / 6.022 X 1023

= 2.94 moles of carbon.

3) 2 KI + Pb(NO3)2⟶PbI2 + 2 KNO3

Mol wt of KI = 166 g/mol and wt of KI = 89.7 g Hence moles of KI = wt / mol wt = 89.7 g / 166 g/mol = 0.540 moles of KI used

2 moles of KI will produce 1 mole of PbI2

then 0.540 moles of KI will produce = 0.540 mol * 1 / 2 = 0.27 moles of PbI2

Mol wt of PbI2 = 461.00 g/mol

= 0.27 mol * 461 g/mol = 124.47 of  PbI2  

Hence 124.47 g of PbI2  are procuced.

4)  2 NaOH + H2SO4 ⟶ Na2SO4 + 2 H2O Mol wt of H2

Mol wt. of NaOH = 40 g/mol and wt of NaOH given = 33.1 g, hence moles of NaOH = 33.1 g / 40 g/mol = 0.828 mol

Mol wt of H2SO4 = 98.08 g/mol

2 moles of NaOH required 1 moles of H2SO4 to react completly.

Then 0.828 moles of NaOH required = 0.828 * 1 / 2 = 0.414 moles of H2SO4

= Wt of H2SO4 = moles of H2SO4   * mol wt. = 0.414 * 98 = 40.57 g H2SO4  

Hence 40.57 g of H2SO4 required to react 33.1 g NaOH