How should the following reaction be classified?

Calcium dihydrogen phosphate, Ca(H₂PO₄)₂, and sodium hydrogen carbonate, NaHCO₃, are ingredients of baking powder that react to

NikAS [45]

0.012 mol of CO₂ can be produced from 3.50 g of baking powder.

<h3>What is baking powder?</h3>

Baking powder is a dry chemical leavener composed of carbonate or bicarbonate and a weak acid.
The addition of a buffer, such as cornstarch, prevents the base and acid from reacting prematurely.
Baking powder is used in baked goods to increase volume and lighten the texture.

To find how many moles of CO₂ are produced from 1.00 g of baking powder:

The balanced equation is:

Ca(H₂PO₄)₂(s) + 2NaHCO₃(s) → 2CO₂(g) + 2H₂O(g) + CaHPO₄(s) + Na₂HPO₄(s)

On 3.50 g of baking power:

mCa(H₂PO₄)₂ = 0.35 × 3.50 = 1.225 g
mNaHCO₃ = 0.31 × 3.50 = 1.085 g

The molar masses are: Ca = 40 g/mol; H = 1 g/mol; P = 31 g/mol; O = 16 g/mol; Na = 23 g/mol; C = 12 g/mol.

So,

Ca(H₂PO₄)₂: 40 + 4 × 1 + 31 + 8 × 16 = 203 g/mol
NaHCO₃: 23 + 1 + 12 + 3 × 16 = 84 g/mol

The number of moles is the mass divided by molar mass, so:

nCa(H₂PO₄)₂ = 1.225/203 = 0.006 mol
nNaHCO₃ = 1.085/84 = 0.0129 mol

First, let's find which reactant is limiting.

Testing for Ca(H₂PO₄)₂, the stoichiometry is:

1 mol of Ca(H₂PO₄)₂ ---------- 2 mol of NaHCO₃
0.006 of Ca(H₂PO₄)₂ -------- x

By a simple direct three rule:

x = 0.012 mol

So, NaHCO₃ is in excess.

The stoichiometry calculus must be done with the limiting reactant, then:

1 mol of Ca(H₂PO₄)₂ ------------- 2 mol of CO₂
0.006 of Ca(H₂PO₄)₂ -------- x

By a simple direct three rule:

x = 0.012 mol of CO₂

Therefore, 0.012 mol of CO₂ can be produced from 3.50 g of baking powder.

Know more about baking powder here:

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The correct question is given below:

Calcium dihydrogen phosphate, Ca(H2PO4)2, and sodium hydrogen carbonate, NaHCO3, are ingredients of baking powder that react with each other to produce CO2, which causes dough or batter to rise: Ca(H2PO4)2(s) + NaHCO3(s) → CO2(g) + H2O(g) + CaHPO4(s) + Na2HPO4(s)[unbalanced] If the baking powder contains 31.0% NaHCO3 and 35.0% Ca(H2PO4)2 by mass: (a) How many moles of CO2 are produced from 3.50 g of baking powder?

3 0

2 years ago

A certain compound is made up of one phosphorus (P) atom, three chlorine (Cl) atoms, and one oxygen (O) atom. What is the chemic

user100 [1]

<span> POCl3 is the correct way to write the chemical formula for this compound</span>

5 0

3 years ago

Limiting Reactants—————-

denis-greek [22]

Answer:

21.8 grams.

Explanation:

Molar mass data from a modern periodic table:

Mg: 24.301;
O: 15.999.

How many moles of MgO will be produced if Mg is the limiting reactant?

Number of moles of Mg:

$\displaystyle n = \frac{m}{M} = \frac{16.3}{24.301} = 0.670644\;\text{mol}$ .

The ratio between the coefficient of Mg and that of MgO is 2:2. Two moles of Mg will make two moles of MgO. 0.670644 moles of MgO will be produced if Mg is the limiting reactant.

How many moles of MgO will be produced if O₂ is the limiting reactant?

Number of moles of O₂:

$\displaystyle n = \frac{m}{M} = \frac{4.33}{15.999} = 0.270642\;\text{mol}$ .

The ratio between the coefficient of O₂ and that of MgO is 1:2. One mole of O₂ will make two moles of MgO. $2\times 0.270642 = 0.541284\;\text{mol}$ of MgO will be produced if O₂ is in excess.