Question

question 10 (1 point)
use the molar mass of ca3(po3)2 to calculate the moles of oxygen in 6.66 x 10^6 g of cp. choose the closest answer.
(6.66 x 10^6 g cp)(-------) =1 mol cp (-------) = 2 mol o
a. 1.112 x 10^10 b. 3.983 x 10^3 c. 2.394 x 10^4 d. 0.03595 e. 2.898
f. 2.162 x 10^22 g. 0.3451 h. 1.675 x 10^25 i. 1.853 x 10^9 j. 34.51
k. 1.436 x 10^5 l. 2.782 x 10^3 m. 3.996 x 10^7 n. 5.969 x 10^-26
o. 5.969 x 10^26 p. 3.595 x 10^-2 q. 2.782 x 10^-3

Explanation:

Step1: Calculate molar mass of \(Ca_3(PO_3)_2\)

Molar mass of \(Ca = 40.08\space g/mol\), \(P = 30.97\space g/mol\), \(O = 16.00\space g/mol\)

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Step2: Calculate moles of \(Ca_3(PO_3)_2\)

Use formula \(n=\frac{m}{M}\), where \(m = 6.66\times10^{6}\space g\), \(M = 278.18\space g/mol\)

$$n_{Ca_3(PO_3)_2}=\frac{6.66\times 10^{6}\space g}{278.18\space g/mol}\approx2.394\times 10^{4}\space mol$$

Step3: Calculate moles of \(O\)

In \(Ca_3(PO_3)_2\), there are 6 \(O\) atoms per formula unit.

$$n_O = 6\times n_{Ca_3(PO_3)_2}=6\times2.394\times 10^{4}\space mol = 1.436\times 10^{5}\space mol$$

Answer:

1. C. \(2.394\times 10^{4}\)
2. K. \(1.436\times 10^{5}\)