1) For the reaction shown, calculate how many grams of oxygen form when each qua

Question

1) For the reaction shown, calculate how many grams of oxygen form when each quantity of reactant completely reacts.
2HgO(s)?2Hg(l)+O2(g). Please show answers in grams.

1) 2.51 gHgO

2) 6.51 gHgO

3) 1.51 kgHgO

4) 3.51 mgHgO

2) For the reaction, calculate how many moles of the product form when 2.53 mol of H2 completely reacts.
Assume that there is more than enough of the other reactant.
H2(g)+Cl2(g)?2HCl(g)

3) For the reaction, calculate how many moles of the product form when 1.75 mol of O2 completely reacts.
Assume that there is more than enough of the other reactant.
2H2(g)+O2(g)?2H2O(l)

4) For the reaction, calculate how many moles of the product form when 1.87 mol of Na completely reacts.
Assume that there is more than enough of the other reactant.
2Na(s)+O2(g)?Na2O2(s)

5) For the reaction, calculate how many moles of the product form when 2.09 mol of O2 completely reacts.
Assume that there is more than enough of the other reactant.
2S(s)+3O2(g)?2SO3(g)

Explanation / Answer

1) 2HgO(s)?2Hg(l)+O2(g)

MOLAR MASS OF HgO =216.6 g/mol

1) 2.51 g HgO

As per equation 216.6 g of HgO yields 32 g of oxygen (1 mole of oxygen).

2.51 g of HgO produces = 2.51 x 32/216.6

= 0.3708 g of oxygen produced.

LET US EVALUATE 32/216.6, FACTOR SO THAT CAN BE MULTIPIED BY GIVEN MASS OF HgO

32/216.6 = 0.14774

2) 6.51 gHgO PRODUCES = 6.51 X 0.14774 g = 0.9618 g of oxygen

3) 1.51 kgHgO PRODUCES = 1.51 x 0.14774 g = 0.2231g 0f oxygen

4) 3.51 mgHgO PRODUCES = 3.51 x 0.14774 = 0.5186 g of oxygen

Q2 )   H2(g)+Cl2(g)?2HCl(g)

1mole of H2 reacts to give = 2 moles of HCl (product)

2.53 mol of H2 reacts completely to give = 2.53 x 2 moles of HCl

= 5.06 moles of HCl

Q3) 2H2(g)+O2(g)?2H2O(l)

1 MOLE OF OXYGEN GIVES 2 MOLES OF WATER (PRODUCT)

1.75 mol of O2 REACTS COMPLETELY TO GIVE = 1.75 X 2 MOLES OF WATER

3.5 MOLES OF H2O (l)

Q4)  2Na(s)+O2(g)?Na2O2(s)

2 mole sof sodium reacts completely to give = 1 mole of sodium peroxide.

1.87 mol of Na completely reacts to give = 1.87/2 moles of product

= 0.935 moles of product sodium peroxide formed.

Q5)   2S(s)+3O2(g)?2SO3(g)

3 MOLES OF OXYGEN GIVES 2 MOLES OF PRODUCT , SO3

2.09 mole of O2 COMPLETETLY REACTS TO GIVE = 2.09 X 2/3 moles of SO3

= 1.3933 MOLES OF SULPHUR TRIOXIDE.

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