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1.The relative atomic mass of metal X is 69.7. 4.53g of X are allowed to react with excess oxygen until completely oxidized. 6.10 g of oxide are obtained. What is the empirical formula of the oxide?

2. P,Q,R and S represent four different compounds. P and Q react according to the following equation: 2P + Q-> R+2S p grams of P react with q grams of Q to give r grams of R and s grams of S. What is the value of s in terms of p, q and r?

3. Rhodonite is a decorative stone that consists of pink MnSiO3. Less valuable variants of the stones have black streaks of MnO2 in them. Analysis of a particular smaple of rhodonite indicates that it contains a total of 49.2% of Mn by mass. What is the percentage by mass of MnO2 in the sample?

### 1 Answer

- andrewLv 710 years agoFavorite Answer
1.The relative atomic mass of metal X is 69.7. 4.53g of X are allowed to react with excess oxygen until completely oxidized. 6.10 g of oxide are obtained. What is the empirical formula of the oxide?

Mole ratio X : O = 4.53/69.7 : (6.10 - 4.53)/16 = 0.065 : 0.0981 = 2 : 3

Empirical formula of the oxide = X2O3

2. P,Q,R and S represent four different compounds. P and Q react according to the following equation: 2P + Q-> R+2S p grams of P react with q grams of Q to give r grams of R and s grams of S. What is the value of s in terms of p, q and r?

By the law of mass conservation:

Total mass of reactants = Total mass of products

p + q = r + s

Hence, s = p + q - r

3. Rhodonite is a decorative stone that consists of pink MnSiO3. Less valuable variants of the stones have black streaks of MnO2 in them. Analysis of a particular smaple of rhodonite indicates that it contains a total of 49.2% of Mn by mass. What is the percentage by mass of MnO2 in the sample?

Mass fraction of Mn in MnSiO3 = 54.9/(54.9 + 28.1 + 16x3) = 54.9/131

Mass fraction of Mn in MnO2 = 54.9 + 16x2 = 54.9/86.9

Let y% be the percentage by mass of MnO2.

(100 - y)%(54.9/131) + y%(54.9/86.9) = 49.2%

(100 - y)(0.4191) + y(0.6318) = 49.2

41.91 - 0.4191y + 0.6318y = 49.2

0.2127y = 7.29

y = 34.3

Hence, the % by mass of MnO2 in the sample = 34.3%

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