Question

For the reaction, calculate how many grams of the product form when 22.4 g of Ca completely reacts.
Assume that there is more than enough of the other reactant.
Ca(s)+Cl2(g)→CaCl2(s)

For the reaction, calculate how many grams of the product form when 22.4 g of Br2 completely reacts.
Assume that there is more than enough of the other reactant.
2K(s)+Br2(l)→2KBr(s)

For the reaction, calculate how many grams of the product form when 22.4 g of O2 completely reacts.
Assume that there is more than enough of the other reactant.
4Cr(s)+3O2(g)→2Cr2O3(s)

For the reaction, calculate how many grams of the product form when 22.4 g of Sr completely reacts.
Assume that there is more than enough of the other reactant.
2Sr(s)+O2(g)→2SrO(s)

Answer 1

Question 1

Ca(s)+Cl2(g) ===> CaCl2(s)

Mass of calcium = 22.4 gm

Molar mass calcium =  40.078 g/mol

Moles of Calcium = 22.4 g / 40.078 g/mol = 0.5589 Moles

Moles of CaCl2 produced = 0.5589 Moles

Mass of CaCl2 produced = 0.5589 Moles x 110.98 g/mol = 62.027 g

Question 2

2K(s)+Br2(l) ====> 2KBr(s)

Mass of Bromine = 22.4 gm

Molar mass calcium = 159.808 g/mol

Moles of Calcium = 22.4 g / 159.808 g/mol = 0.1401 Moles

Moles of KBr produced = 0.2803 Moles

Mass of KBr produced = 0.2803 Moles x 119.002 g/mol = 33.36 g

Question 3

4Cr(s)+3O2(g) ===> 2Cr2O3(s)

Mass of oxygen = 22.4 gm

Molar mass of oxygen = 32 g/mol

Moles of oxygen = 22.4 g / 32 g/mol = 0.700 Moles

Moles of Cr2O3 produced = 0.4666 Moles

Mass of Cr2O3 produced = 0.4666 Moles x 151.99 g/mol = 70.931 g

Question 3

2Sr(s)+O2(g) ==> 2SrO(s)

Mass of Sr = 22.4 gm

Molar mass of Sr = 87.62 g/mol

Moles of Sr = 22.4 g / 87.62 g/mol = 0.2556 Moles

Moles of SrO produced = 0.2556 Moles

Mass of SrO produced = 0.2556 Moles x 103.61 g/mol = 26.49 g