Given the following reaction:

3 CaCl2 + 2 H3PO4 Ca3(PO4)2 + 6 HCl
How many grams of H3PO4 are needed to react completely with 1.837 g of CaCl2?

I don't understand at all

cmon, we just did one of these. convert grams to moles, then use the reaction to figure the relative amounts needed/produced.

moles of input = 1.837/111 = .01655 moles CaCl2

Each 3 CaCl2 requires 2 H3PO4

so, 2/3 * .01655 = .011 moles H3PO4

.011 moles H3PO4 = .011 * 98 = 1.081g

3 CaCl2 + 2 H3PO4 Ca3(PO4)2 + 6 HCl

How many moles of HCl will be produced if 3.525 g of H3PO4 are reacted completely?

H3POA(1) + CaCh(aq) =Cag(POA)2(s) + HC1 (ag)

A. How many moles of calcium chloride are required to react to produce 0.873 moles
of calcium phosphate?
B. How many moles of phosphoric acid are required to react with 2.14 moles of
calcium chloride?

To solve this problem, you need to use stoichiometry, which is a mathematical approach to determine the amount of reactants and products in a chemical reaction. Here's how you can calculate the grams of H3PO4 needed to react completely with 1.837 g of CaCl2:

1. Start by finding the molar mass of CaCl2 and H3PO4. The molar masses can be found on the periodic table:
Molar mass of CaCl2 = 40.08 g/mol + 2 * 35.45 g/mol = 110.98 g/mol
Molar mass of H3PO4 = 1 * 1.01 g/mol + 3 * 1.01 g/mol + 4 * 16.00 g/mol = 97.99 g/mol

2. Convert the given mass of CaCl2 to moles. Use the formula:
Moles = Mass / Molar mass
Moles of CaCl2 = 1.837 g / 110.98 g/mol ≈ 0.0165 mol

3. Use the balanced equation to determine the stoichiometric ratio between CaCl2 and H3PO4. From the equation:
3 CaCl2 + 2 H3PO4 → Ca3(PO4)2 + 6 HCl
The ratio is 3:2.

4. Calculate the number of moles of H3PO4 needed to react with the calculated moles of CaCl2. Use the ratio from step 3:
Moles of H3PO4 = (0.0165 mol CaCl2) * (2 mol H3PO4 / 3 mol CaCl2) ≈ 0.0110 mol

5. Finally, convert the moles of H3PO4 to grams:
Mass of H3PO4 = Moles * Molar mass
Mass of H3PO4 = 0.0110 mol * 97.99 g/mol ≈ 1.08 g

Therefore, approximately 1.08 grams of H3PO4 are needed to react completely with 1.837 grams of CaCl2.