1)The following balanced chemical equation represents the reaction of calcium carbonate CaCO3 (s), with hydrochloric acid :

CaCo3(s) + 2HCl (aq) -> CaCl2(aq) + CO2(g) + H2O(l)
If 155 g of calcium carbonate reacts with 245 g of hydrochloric acid, what mass of calcium chloride, CaCl2(s), is produced?

2). The reaction between solid white phosphorus, P4(s), and oxygen gas produces solid tetraphosphorus decoxide that is formed when 25.0g of solid white phosphorus and 50.0 g of oxygen are combined.

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mols CaCo3 = grams/molar mass.

mols HCl = grams/molar mass.

Convert mols CaCO3 to mols CaCl2 using the coefficients in the balanced equation.
Do the same for mols HCl.
It is likely that these two values for mols CaCl2 will not be the same which means one is not right; the correct one in limiting reagents (this is a limiting reagent problem) is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.

Using the smaller value convert mols CaCl2 to grams. g = mols x molar mass.

P4 is the same thing; however, you didn't ask a question for #2.

Sure! To find the mass of calcium chloride produced in the reaction, we need to use the concept of stoichiometry. Stoichiometry is a branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions.

Step 1: Calculate the number of moles of reactants.

First, we need to convert the given masses of reactants (calcium carbonate and hydrochloric acid) into moles by using their respective molar masses.

The molar mass of calcium carbonate (CaCO3) can be calculated by adding up the atomic masses of its constituent elements:

Ca: 40.08 g/mol
C: 12.01 g/mol
O: 16.00 g/mol (3 oxygen atoms in CaCO3)

Molar mass of CaCO3 = (40.08 g/mol) + (12.01 g/mol) + (16.00 g/mol x 3) = 100.09 g/mol

To find the number of moles of calcium carbonate, divide the given mass (155 g) by its molar mass:

Number of moles of CaCO3 = 155 g / 100.09 g/mol = 1.549 mol

Similarly, the molar mass of hydrochloric acid (HCl) is:

H: 1.01 g/mol
Cl: 35.45 g/mol

Molar mass of HCl = 1.01 g/mol + 35.45 g/mol = 36.46 g/mol

Number of moles of HCl = 245 g / 36.46 g/mol = 6.71 mol (rounded to two decimal places)

Step 2: Determine the stoichiometric ratio of reactants and products.

From the balanced equation:
CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l)

We can see that the stoichiometric ratio between CaCO3 and CaCl2 is 1:1. Hence, for every 1 mole of calcium carbonate, 1 mole of calcium chloride is produced.

Step 3: Calculate the mass of calcium chloride produced.

Since the number of moles of calcium carbonate and calcium chloride is the same (1.549 mol), we can calculate the mass of calcium chloride produced by multiplying the number of moles by its molar mass:

Molar mass of CaCl2 = 40.08 g/mol (Ca) + 2(35.45 g/mol) (Cl) = 110.98 g/mol

Mass of CaCl2 = 1.549 mol x 110.98 g/mol = 171.57 g

Therefore, the mass of calcium chloride produced in the reaction is 171.57 grams.

Moving on to the second question:

Step 1: Convert the given masses of reactants (solid white phosphorus and oxygen) into moles.

The molar mass of white phosphorus (P4) is the sum of the atomic masses of phosphorus (P):

Molar mass of P4 = 4(30.97 g/mol) = 123.88 g/mol

Number of moles of white phosphorus = 25.0 g / 123.88 g/mol = 0.202 mol (rounded to three decimal places)

The molar mass of oxygen (O2) is:

O: 16.00 g/mol (2 oxygen atoms in O2)

Molar mass of O2 = 2(16.00 g/mol) = 32.00 g/mol

Number of moles of oxygen = 50.0 g / 32.00 g/mol = 1.563 mol (rounded to three decimal places)

Step 2: Determine the stoichiometric ratio of reactants and products.

From the balanced equation:
P4(s) + 5O2(g) -> P4O10(s)

We can see that the stoichiometric ratio between P4 and P4O10 is 1:1. Hence, for every 1 mole of white phosphorus, 1 mole of tetraphosphorus decoxide is produced.

Step 3: Calculate the mass of tetraphosphorus decoxide produced.

Since the number of moles of white phosphorus and tetraphosphorus decoxide is the same (0.202 mol), we can calculate the mass of tetraphosphorus decoxide by multiplying the number of moles by its molar mass:

Molar mass of P4O10 = 4(30.97 g/mol) + 10(16.00 g/mol) = 283.89 g/mol

Mass of P4O10 = 0.202 mol x 283.89 g/mol = 57.36 g (rounded to two decimal places)

Therefore, the mass of tetraphosphorus decoxide produced in the reaction is 57.36 grams.