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VladimirAG [237]
3 years ago
5

Calcium phosphate reacts with hydrochloride acid to produce calcium chloride and phosphoric acid. This is the balanced chemical

equation for this reaction.
6HCl + Ca3(PO4)2 -> 3CACl2 + 2H3PO4.
What mass of phosphoric acid can be produced from 103 grams of calcium phosphate?

Answer: 103 grams of calcium phosphate can produce_____ moles and _____grams of phosphoric acid.
Chemistry
1 answer:
Mashcka [7]3 years ago
6 0

Answer: 0.664 moles and 65.1 grams EDMENTUM USERS

Explanation: From the equation, we know that 1 mole of calcium phosphate produces 2 moles of phosphoric acid (H3PO4). So, if we know how many moles of calcium phosphate are present in 103 grams of Ca3(PO4)2, we can find the corresponding number of moles of H3PO4 that are produced during the reaction.

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marishachu [46]

Answer:

(a) -0.00017 M/s;

(b) 0.00034 M/s

Explanation:

(a) Rate of a reaction is defined as change in molarity in a unit time, that is:

r = \frac{\Delta c}{\Delta t}

Given the following reaction:

2 N_2O_5 (g)\rightleftharpoons 4 NO_2 (g) + O_2 (g)

We may write the rate expression in terms of reactants firstly. Since reactants are decreasing in molarity, we're adding a negative sign. Similarly, if we wish to look at the overall reaction rate, we need to divide by stoichiometric coefficients:

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Reaction rate is also equal to the rate of formation of products divided by their coefficients:

r = \frac{\Delta [NO_2]}{4 \Delta t} = \frac{\Delta [O_2]}{\Delta t}

Let's find the rate of disappearance of the reactant firstly. This would be found dividing the change in molarity by the change in time:

r_{N_2O_5} = \frac{0.066 M - 0.100 M}{200.00 s - 0.00 s} = -0.00017 M/s

(b) Using the relationship derived previously, we know that:

-\frac{\Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{4 \Delta t}

Rate of appearance of nitrogen dioxide is given by:

r_{NO_2} = \frac{\Delta [NO_2]}{\Delta t}

Which is obtained from the equation:

-\frac{\Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{4 \Delta t}

If we multiply both sides by 4, that is:

-\frac{4 \Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{\Delta t}

This yields:

[tex]r_{NO_2} = \frac{\Delta [NO_2]}{\Delta t} = -2\frac{\Delta [N_2O_5]}{ \Delta t} = -2\cdot (-0.00017 M/s) = 0.00034 M/s[tex]

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