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2.8 g of CaCl2 is needed to react with 2.50 g of sodium carbonate.

Explanation:

  • The balanced equation is given as

              Na2CO3 +CaCl2  ----> 2 Nacl + CaCO3

  • Use the balanced equation to determine the mole ratios between  CaCO 3  and  CaCl 2 .

Calcium carbonate and calcium chloride.

                      [tex]\frac{1 mol CaCO3}{1 mol CaCl2}[/tex] [tex]and[/tex] [tex]\frac{1 mol CaCl2}{1 mol CaCO3}[/tex]

  • Determine the moles of each reactant by dividing the given masses by their molar masses.

2.50 g of CaCO3 [tex]\times[/tex] (1 mol of CaCO3 / 100.1 g of CaCO3) = 0.025 g of CaCO3.

  • Determine the mass of  CaCl 2  produced by each reactant by multiplying the moles of each reactant times the mole ratios with  CaCl 2  in the numerator. Then multiply the result by the molar mass of  CaCl 2  ( 111 g/mol ) .

0.025 g of CaCO3 [tex]\times[/tex] (1 mol of CaCl2 / 1 mol of CaCO3) [tex]\times[/tex] (111 g of CaCl2 / 1 mol of CaCl2)  = 2.8 g of CaCl2.

mahima6824soni

2.8 grams of [tex]\bold{CaCl_2}[/tex] is needed to react with 2.50 grams of sodium carbonate.

What is mass?

  • Mass is the quantity of matter or average mass of the atoms of an element.

Calculating the mass of [tex]\bold{CaCl_2}[/tex]

Given equation, [tex]\bold{Na_2CO_3 +CaCl_2 = 2 Nacl + CaCO_3}[/tex]

Given the quantity of sodium is 2.50 grams.

[tex]\bold{2.50g\;of\;CaCO3 \;\dfrac{(1 mol\; of\; CaCO_3)}{(100.1\; g\; of\; CaCO3)} = 0.025 g\; of \;CaCO3}[/tex]

  • Now, determining the mass of [tex]\bold{CaCl_2}[/tex] produced, then multiply the result by the molar mass of [tex]\bold{CaCl_2}[/tex].

[tex]\bold{0.025 g\; of\; CaCO3\;\dfrac{1\; mol\; of \;CaCl2}{1 mol\; of\; CaCO3}\times\dfrac{111\; g\; of\; CaCl2}{1\; mol\; of \;CaCl2} = 2.8 g\; of \;CaCl2}[/tex]

Thus, the  mass of calcium chloride needed is 2.8 grams.

Learn more about calcium carbonate here:

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