To solve this problem, we need to use stoichiometry, which is a method in chemistry that uses the relationships between reactants and products in a chemical reaction to determine desired quantitative data.

Here are the steps:

1. First, we need to find the molar mass of Ca3(PO4)2 and (NH4)3PO4.

The molar mass of Ca3(PO4)2 is approximately 310.18 g/mol.
The molar mass of (NH4)3PO4 is approximately 149.09 g/mol.

2. Next, we need to convert the given mass of Ca3(PO4)2 to moles. We do this by dividing the given mass by the molar mass.

Moles of Ca3(PO4)2 = 600 g / 310.18 g/mol = 1.934 moles

3. Now, we use the stoichiometric ratio from the balanced chemical equation to find the moles of (NH4)3PO4 needed. From the balanced equation, we can see that 2 moles of (NH4)3PO4 produce 1 mole of Ca3(PO4)2.

Moles of (NH4)3PO4 = 1.934 moles Ca3(PO4)2 * (2 moles (NH4)3PO4 / 1 mole Ca3(PO4)2) = 3.868 moles (NH4)3PO4

4. Finally, we convert the moles of (NH4)3PO4 to grams by multiplying by the molar mass.

Mass of (NH4)3PO4 = 3.868 moles * 149.09 g/mol = 576.6 g

So, approximately 576.6 g of (NH4)3PO4 is required to generate 600 g of Ca3(PO4)2.

Question

To solve this problem, we need to use stoichiometry, which is a method in chemistry that uses the relationships between reactants and products in a chemical reaction to determine desired quantitative data.

Here are the steps:

1. First, we need to find the molar mass of Ca3(PO4)2 and (NH4)3PO4.

The molar mass of Ca3(PO4)2 is approximately 310.18 g/mol.
   The molar mass of (NH4)3PO4 is approximately 149.09 g/mol.

2. Next, we need to convert the given mass of Ca3(PO4)2 to moles. We do this by dividing the given mass by the molar mass.

Moles of Ca3(PO4)2 = 600 g / 310.18 g/mol = 1.934 moles

3. Now, we use the stoichiometric ratio from the balanced chemical equation to find the moles of (NH4)3PO4 needed. From the balanced equation, we can see that 2 moles of (NH4)3PO4 produce 1 mole of Ca3(PO4)2.

Moles of (NH4)3PO4 = 1.934 moles Ca3(PO4)2 * (2 moles (NH4)3PO4 / 1 mole Ca3(PO4)2) = 3.868 moles (NH4)3PO4

4. Finally, we convert the moles of (NH4)3PO4 to grams by multiplying by the molar mass.

Mass of (NH4)3PO4 = 3.868 moles * 149.09 g/mol = 576.6 g

So, approximately 576.6 g of (NH4)3PO4 is required to generate 600 g of Ca3(PO4)2.

Knowee AI · Accepted Answer

To solve this problem, we need to use stoichiometry, which is a method in chemistry that uses the relationships between reactants and products in a chemical reaction to determine desired quantitative data.

Here are the steps:

1. First, we need to find the molar mass of Ca3(PO4)2 and (NH4)3PO4.

The molar mass of Ca3(PO4)2 is approximately 310.18 g/mol.
   The molar mass of (NH4)3PO4 is approximately 149.09 g/mol.

2. Next, we need to convert the given mass of Ca3(PO4)2 to moles. We do this by dividing the given mass by the molar mass.

Moles of Ca3(PO4)2 = 600 g / 310.18 g/mol = 1.934 moles

3. Now, we use the stoichiometric ratio from the balanced chemical equation to find the moles of (NH4)3PO4 needed. From the balanced equation, we can see that 2 moles of (NH4)3PO4 produce 1 mole of Ca3(PO4)2.

Moles of (NH4)3PO4 = 1.934 moles Ca3(PO4)2 * (2 moles (NH4)3PO4 / 1 mole Ca3(PO4)2) = 3.868 moles (NH4)3PO4

4. Finally, we convert the moles of (NH4)3PO4 to grams by multiplying by the molar mass.

Mass of (NH4)3PO4 = 3.868 moles * 149.09 g/mol = 576.6 g

So, approximately 576.6 g of (NH4)3PO4 is required to generate 600 g of Ca3(PO4)2.

You want to create 600. g Ca3(PO4)2 and need to know how much of each reactant to use.2(NH4)3PO4 + 3Ca(C2H3O2)2 → 6NH4C2H3O2 + Ca3(PO4)2(S)What mass of (NH4)3PO4 is required to generate 600. g Ca3(PO4)2?[?]g (NH4)3PO4

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